**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
import sys
import threading

def main():
    import sys

    sys.setrecursionlimit(1 << 25)
    t = int(sys.stdin.readline())
    target = ['n', 'a', 'r', 'e', 'k']

    for _ in range(t):
        n, m = map(int, sys.stdin.readline().split())
        strings = [sys.stdin.readline().strip() for _ in range(n)]

        # Initialize DP
        prev_dp = [-float('inf')] * 5
        prev_dp[0] = 0

        for s in range(5):
            pass  # Just to ensure s is from 0 to4

        for s_idx, s in enumerate(range(n)):
            # Process each string
            # To save memory, process one string at a time
            # But here, strings are stored
            # Precompute for each state
            # Initialize a list to store the precomputed transitions
            # For each string, precompute for each s=0 to4
            # However, since we process strings sequentially, we can process them on the fly

            # Not storing all precomputations, process on the fly
            pass

        # Reset the strings index
        # Now process each string
        prev_dp = [-float('inf')] * 5
        prev_dp[0] = 0

        for string in strings:
            # Precompute for each s=0 to4
            transitions = []
            for s_start in range(5):
                s_curr = s_start
                k_incr =0
                C_i =0
                for c in string:
                    if c in target:
                        C_i +=1
                        if c == target[s_curr]:
                            s_curr +=1
                            if s_curr ==5:
                                k_incr +=1
                                s_curr =0
                transitions.append( (s_curr, k_incr, C_i) )
            # Now update DP
            next_dp = prev_dp.copy()
            for s_start in range(5):
                if prev_dp[s_start] != -float('inf'):
                    s_end, delta_k, C_i = transitions[s_start]
                    new_val = prev_dp[s_start] +10*delta_k - C_i
                    if new_val > next_dp[s_end]:
                        next_dp[s_end] = new_val
            prev_dp = next_dp

        # After all strings, compute max(prev_dp[s] -s) for s=0 to4
        max_val = -float('inf')
        for s in range(5):
            current = prev_dp[s] - s
            if current > max_val:
                max_val = current
        answer = max(0, max_val)
        print(int(answer))

threading.Thread(target=main,).start()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider scenarios where the algorithm might fail to correctly compute the optimal score. Here are some key points to consider:

1. **Edge Cases:**
   - Strings that do not contain all the required letters ('n', 'a', 'r', 'e', 'k').
   - Strings that contain some but not all of the required letters.
   - Strings that contain all the required letters but in a way that Narek cannot form a complete "narek" sequence.

2. **Complex Scenarios:**
   - Strings that contain multiple occurrences of the required letters.
   - Strings that contain letters in a way that Narek can form multiple "narek" sequences but with some leftover letters.

3. **Optimal Selection:**
   - Ensure that the test case requires careful selection of strings to maximize the score.

Given these considerations, let's create a test input that includes a mix of these scenarios:

```plaintext
3
5 5
narek
narek
narek
narek
narek
5 5
narek
narek
narek
narek
zzzzz
5 5
narek
narek
narek
zzzzz
zzzzz
```

### Explanation:

1. **First Test Case:**
   - All strings are "narek".
   - Narek can select all strings and form 5 "narek" sequences.
   - Narek's score: \(5 \times 5 = 25\).
   - ChatGPT's score: \(0\) (since all letters are used).
   - Final score: \(25 - 0 = 25\).

2. **Second Test Case:**
   - Four strings are "narek" and one string is "zzzzz".
   - Narek can select the four "narek" strings and form 4 "narek" sequences.
   - Narek's score: \(4 \times 5 = 20\).
   - ChatGPT's score: \(0\) (since all letters in the "narek" strings are used).
   - Final score: \(20 - 0 = 20\).

3. **Third Test Case:**
   - Three strings are "narek" and two strings are "zzzzz".
   - Narek can select the three "narek" strings and form 3 "narek" sequences.
   - Narek's score: \(3 \times 5 = 15\).
   - ChatGPT's score: \(0\) (since all letters in the "narek" strings are used).
   - Final score: \(15 - 0 = 15\).

This test input should expose any issues in the buggy code, particularly in how it handles the selection and scoring of strings.

3
5 5
narek
narek
narek
narek
narek
5 5
narek
narek
narek
narek
zzzzz
5 5
narek
narek
narek
zzzzz
zzzzz


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
import java.util.*;
public class Main
{
    static boolean cnt(char c, char word[]){
        for(char cc : word)
            if(cc == c)
                return true;
        return false;
    }
    static int func(int i,int j,String arr[], char word[], int dp[][]){
        if(i>=arr.length)
            return 0;
        if(dp[i][j] != -1)
            return dp[i][j];
        int val = 0, t = j;
        for(int k = 0;k<arr[i].length();k++){
            if(arr[i].charAt(k) == word[j]){
                if(j == 4)
                    val += 5;
                j = (j+1)%5;
            }else if(cnt(arr[i].charAt(k), word))
                val--;
        }
        if(i == arr.length-1)
            val -= j;
        val += func(i+1, j, arr, word, dp);
        return dp[i][t] = Math.max(val, func(i+1, t, arr, word, dp));
    }
    public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int tt = sc.nextInt();
		while(tt-->0){
		    int n = sc.nextInt(), m =sc.nextInt();
		    sc.nextLine();
		    String arr[] = new String[n];
		    char word[] = {'n','a','r','e','k'};
		    for(int i = 0;i<n;i++){
		        arr[i] = sc.nextLine();
		    }
		    int dp[][] = new int[n][5];
		    for(int i =0;i<n;i++)
		        Arrays.fill(dp[i], -1);
	        System.out.println(func(0,0,arr, word, dp));
		}
	}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5 5
narek
narek
narek
narek
narek
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
import java.util.Arrays;
import java.util.Scanner;

public class problemC {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String narek = "narek";
    int tests = sc.nextInt();
    while (tests-- > 0) {
      int n = sc.nextInt();
      int m = sc.nextInt();
      int[][] dp = new int[n + 1][5];
      for (int[] arr : dp) {
        Arrays.fill(arr, (int) -1e9);
      }
      dp[0][0] = 0;
      for (int i = 1; i <= n; i++) {
        String s = sc.next();
        for (int j = 0; j < 5; j++) {
          dp[i][j] = dp[i - 1][j];// Score without considering current string with prefixes (j=0)"",(j=1)"n",(j=2)"na",(j=3)"nar",(j=4)"nare"
          int ind = j;// index for elem considered (ind=0)"n",(ind=1)"a",(ind=2)"r",(ind=3)"e",(ind=4)"k"
          int score = 0;
          for (int k = 0; k < m; k++) {
            if (narek.indexOf(s.charAt(k)) != -1) {
              score--;
            }
            if (narek.charAt(ind) == s.charAt(k)) {
              ind++;
            }
            if (ind == 5) {
              //              If all chars found restart the sequence
              score += 10;
              ind = 0;
            }
          }
          dp[i][j] = Math.max(dp[i][j], dp[i][j] + score);
        }
      }
      System.out.println(Arrays.stream(dp[n]).max().getAsInt());
    }
  }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
```

4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define scn(v) for(auto &i:v)cin>>i

ll n,m;
vector<string> v;
string nrk="narek";
map<char,int> fre;
ll dp[10000][50];
ll vis[10000][50];
ll txt=50;
ll goDP(int idx,int p){
    if(idx==n){
        return -p;
    }
    ll& ret = dp[idx][p];
    if(vis[idx][p]==txt){
        return ret;
    }
    vis[idx][p]=txt;

    ll ch1= goDP(idx+1,p);
    ll ch2=0;
    ret=0;
    int ind=p;

        for(char x : v[idx]){
            if(x==nrk[ind]){
                ind++;
                if(ind==5){
                    ch2+=5;
                    ind=0;
                }
            }else if(fre[x]){
                ch2--;
            }
        }

    ch2 += goDP(idx + 1, ind);
    ret= max({ch1, ch2,0ll});

    return ret;
}

void solve(){

    cin>>n>>m;
    v=vector<string>(n);
    scn(v);

txt++;

    cout<<goDP(0,0);

}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cin.tie(NULL);

    fre['n']++;
    fre['a']++;
    fre['r']++;
    fre['e']++;
    fre['k']++;
    int t = 1;
    cin >> t;

    int tx=1;
    for(int txt=0;txt<t;txt++){
        solve();
        cout<<'\n';
    }

    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
#include <vector>
#include <set>
#include <map>
#include "iostream"
#include "algorithm"
using namespace std;

int t, n, m;
vector<char> narek {'n', 'a', 'r', 'e', 'k'};

void main2() {
    cin >> n >> m;

    vector <long long> count1, count2;
    count1.resize(5);
    count2.resize(5);
    count1[0] = 0;
    count1[1] = count1[2] = count1[3] = count1[4] = LONG_LONG_MIN / 2;

    for (int i = 0; i < n; ++i) {
        std::copy(count1.begin(), count1.end(), count2.begin());

        string s;
        cin >> s;
        for (int k = 0; k < 5; ++k) {
            int c = k;
            long long sum = count1[c];
            for (int j = 0; j < m; ++j) {
                if (s[j] == 'n' ||s[j] == 'a' ||s[j] == 'r' ||s[j] == 'e' ||s[j] == 'k') {
                    --sum;
                }
                if (s[j] == narek[c]) {
                    ++c;
                    if (c == 5) {
                        c = 0;
                        sum += 10;
                    }
                }
            }
            count2[c] = max(count2[c], sum);
        }
//        for (auto x : count1) {
//            cout << x << ' ';
//        }
//        cout << '\n';

        std::copy(count2.begin(), count2.end(), count1.begin());
    }

    cout << count1[0] << "\n";
}

int main() {
    cin >> t;
    for (int i = 0; i < t; ++i) {
        main2();
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
import sys
import math, heapq, random
from collections import defaultdict, Counter, deque
from bisect import bisect_left, bisect_right
def II(): return int(sys.stdin.readline().strip())
def IL(): return list(map(int, sys.stdin.readline().strip().split()))
def S(): return sys.stdin.readline().strip()

def solve():
    n, m = IL()
    s = [S() for _ in range(n)]
    narek = 'narek'
    dp = [0] * 5
    for i in range(n):
        dp2 = dp[:]
        for start in range(5):
            cur = start
            cost = 0
            for j in range(m):
                if s[i][j] == narek[cur % 5]:
                    cost += 1
                    cur += 1
                elif s[i][j] in narek:
                    cost -= 1
            end = cur % 5
            dp2[end - 1] = max(dp2[end - 1], dp[start - 1] + cost)
        dp = dp2
    print(dp[-1])

for _ in range(II()):
    solve()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5 5
narek
naerk
nakre
nkaer
nkera
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
def scoreCounter(word, letter, last_word = False):
    chars = ['n', 'a', 'r', 'e', 'k']
    char_pointer = chars.index(letter)
    scoren = 0
    scorec = 0
    for char in word:
        if char in chars:
            if char == chars[char_pointer]:
                if char == 'k':
                    char_pointer = 0
                    scoren += 5
                else:
                    char_pointer += 1
            else:
                scorec += 1
    if last_word:
        scorec += char_pointer
    return([scoren-scorec, chars[char_pointer]])

t = int(input())
results = []
for _ in range(t):
    n, m = list(map(int, input().split()))
    words = []
    chars = ['n', 'a', 'r', 'e', 'k']
    for _n in range(n):
        words.append(input())
    attributes = []
    for word in words:
        last_word = False
        if words.index(word) == len(word) - 1:
            last_word = True
        for char in chars:
            attributes.append(scoreCounter(word, char, last_word))

    score = dict()
    for i in range(5):
        score[str(n-1)+chars[i]] = max(attributes[-5 + i][0], 0)

    for i in reversed(range(1, n)):
        for chari in range(len(chars)):
            wrdscore, endchar = attributes[(i-1)*5 + chari]
            enchari = chars.index(endchar)
            score[str(i-1)+ chars[chari]] = max(wrdscore + score[str(i)+ endchar], score[str(i)+ chars[chari]])
    results.append(score['0n'])
for result in results:
    print(result)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
5 1
n
a
r
e
k
1 5
narek
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
```

3
5 1
n
a
r
e
k
1 5
narek
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
import java.io.*;
import java.util.*;

public class C {

    private static final String NAME = "narek";

    private static final int[][] dp;

    static {
        dp = new int[1001][5];
        for(int i = 0; i < 1001; i++) {
            Arrays.fill(dp[i], Integer.MIN_VALUE);
        }
    }

    private static int dfs(int i, int j, String[] words) {

        if ( i == words.length) {
            return 0;
        }

        if (dp[i][j] != Integer.MIN_VALUE) return dp[i][j];

        int val = dfs(i + 1, j, words);

        int chatgpt = 0, loops = 0;

        for(int k = 0; k < words[i].length(); k++) {
            char ch = words[i].charAt(k);
            if(NAME.charAt(j) == ch) {
                j = (j + 1)%5;
                if (j == 0) loops++;
            } else if(NAME.contains(String.valueOf(ch))) {
                chatgpt--;
            }
        }

        int narek = loops*5;

        return dp[i][j] =  Math.max(val, chatgpt + narek + dfs(i + 1, j, words));
    }

    public static void main(String[] args) throws IOException {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());

        while(t-- > 0) {
            int[] l = Arrays.stream(br.readLine().split("\\s")).mapToInt(Integer::parseInt).toArray();

            String[] words = new String[l[0]];

            for(int i = 0; i < l[0]; i++) words[i] = br.readLine();

            for(int i = 0; i < l[0]; i++) Arrays.fill(dp[i], Integer.MIN_VALUE);

            System.out.println(dfs(0, 0, words));
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5 2
na
re
nk
ar
ke
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();

        while (t-- > 0) {
            int n = scan.nextInt();
            int m = scan.nextInt();

            char[] s = {'n', 'a', 'r', 'e', 'k'};
            HashSet<Character> set = new HashSet<>();
            for (char item : s) set.add(item);

            char[][] c = new char[n][];
            HashMap<Integer, Integer>[] maps = new HashMap[n];
            for (int i = 0; i < n; i++) {
                c[i] = scan.next().toCharArray();
                maps[i] = new HashMap<>();
                maps[i].put(i, 0);
            }
            int ans = 0;
            for (int i = 0; i < n; i++) {
                for (int time : maps[i].keySet()) {
                    int j = time;
                    int add = 0, dec = 0;
                    for (int k = 0; k < m; k++) {
                        if (c[i][k] == s[j]) {
                            j = (j + 1) % 5;
                            add++;
                        } else {
                            if (set.contains(c[i][k])) dec++;
                        }
                    }
                    maps[i].put(time, maps[i].getOrDefault(time, 0) + (add - dec));
                    for (int k = i + 1; k < n; k++) {
                        if (!maps[k].containsKey(j) || maps[i].get(time) >= maps[k].get(j)) {
                            maps[k].put(j, maps[i].get(time));
                        }
                    }
                    ans = Math.max(maps[i].get(time) - j * 2, ans);
                }
            }
            System.out.println(ans);
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5 5
narek
narek
narek
narek
narek
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
def II():
    return(int(input()))
def LMI():
    return(list(map(int,input().split())))
def I():
    return(input())
def MII():
    return(map(int,input().split()))
import sys
input=sys.stdin.readline
# import io,os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# int(math.log(len(L)))
# import math
# from collections import defaultdict
# mod=10**9+7
# from collections import deque
# import  math

import bisect
# from functools import cache
# @cache
def t():
    n,m=MII()
    ans=0

    for _ in range(n):
        s=I()[:-1]
        V=list("narek")
        poi=0

        temp=0
        for j in range(m):
            if s[j]==V[poi]:
                poi+=1
            elif s[j] in V:
                temp-=1
            if poi==5:
                temp+=5
                poi=0
        ans+=max(temp,0)
    print(ans)

if __name__=="__main__":

    for _ in range(II()):
        t()
    # t()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
```

### Explanation:

1. **Test Case 1:**
   - **Input:** `5 2`
   - **Strings:** `nn`, `aa`, `rr`, `ee`, `kk`
   - **Expected Output:** `0`
   - **Reason:** If Narek selects all strings, the combined string is "nnaarreekk". Narek can find one complete "narek" sequence, scoring 5 points. However, the second sequence of "narek" is also present, which ChatGPT will score, resulting in a score of 5 for ChatGPT. Thus, the final score is `5 - 5 = 0`.

2. **Test Case 2:**
   - **Input:** `1 5`
   - **String:** `narek`
   - **Expected Output:** `5`
   - **Reason:** Narek can find one complete "narek" sequence, scoring 5 points. ChatGPT will not score any points since Narek used all letters. Thus, the final score is `5 - 0 = 5`.

3. **Test Case 3:**
   - **Input:** `1 4`
   - **String:** `nare`
   - **Expected Output:** `0`
   - **Reason:** Narek cannot find the last 'k', so his score remains 0. ChatGPT will score 4 points for the unused letters. Thus, the final score is `0 - 4 = -4`.

4. **Test Case 4:**
   - **Input:** `5 7`
   - **Strings:** `nrrarek`, `nrnekan`, `uuuuuuu`, `ppppppp`, `nkarekz`
   - **Expected Output:** `7`
   - **Reason:** Narek can select the first and the last strings, resulting in the combined string "nrrareknkarekz". Narek can find two complete "narek" sequences, scoring 10 points. The remaining letters that Narek did not use are "r", "z", and "k" (from the second "narek"), which ChatGPT will score, resulting in a score of 3. Thus, the final score is `10 - 3 = 7`.

This test input is designed to cover various edge cases and scenarios to ensure the algorithm handles them correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define mod 1000000007
#define PI 3.1415926535897932384626433
#define mxe(v)  *max_element(v.begin(), v.end())
#define all(v)  v.begin(), v.end()
#define allr(v) v.rbegin(),v.rend()
#define PB push_back
#define F first
#define S second
#define pi pair<ll,ll>
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define derar ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
//for(auto it:v1)
//cout.precision(10);

using namespace std;
const int N = 1e3 + 10;
const int intinf = 1e9 + 10;
const ll llinf = 1e18 + 10;
const ld eps = 1e-9;

string ScanString()
{
    char ch[100000];
    scanf("%s", ch);
    return ch;
}
string s[N];
string f="narek",all;
int n,m;
ll ans[N][7];
bool vis[N][7];
ll dp(int i,int l)
{
    if(i==n)
        return 0;
    if(vis[i][l]==1)
        return ans[i][l];
    ll an=dp(i+1,l);
    int c=0;
    for(int j=0;j<m;j++)
    {
        if(s[i][j]=='n' || s[i][j]=='a' || s[i][j]=='r' || s[i][j]=='e' || s[i][j]=='k')
            c--;
    }
    for(int j=0;j<m;j++)
    {
        if(f[l]==s[i][j])
            l++;
        if(l==f.size())
            c+=10,l=0;
    }
    an=max(an,dp(i+1,l)+c);
    return ans[i][l]=an;
}
void _clear()
{
    for(int i=0;i<n;i++)
        for(int j=0;j<7;j++)
            ans[i][j]=0,vis[i][j]=0;
}
void solve(int test)
{
    cin>>n>>m;
    _clear();
    for(int i=0;i<n;i++)
        cin>>s[i];
    cout<<dp(0,0)<<endl;
}
int main()
{
    derar
    //freopen("window.in","r",stdin);
    int t=1;
    cin>>t;
    for(int i=1; i<=t; i++)
    {
        _clear();
        solve(i);
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
5 5
narek
narek
narek
narek
narek
1 5
narex
2 5
narek
narek
```

3
5 5
narek
narek
narek
narek
narek
1 5
narex
2 5
narek
narek


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
#include <bits/stdc++.h>

#define sz(x) ((int)(x.size()))
#define all(x) x.begin(), x.end()

using namespace std;

typedef long long ll;

vector<string> v;
ll n,m;
const string sp = "narek";
set<char> sps = {'n', 'a', 'r', 'e', 'k'};
bool do1[1010][6];
pair<ll,bool> dp[1010][6];
ll tt1;

pair<ll,int> inteScore(int i, int pr) {
    int cc = pr;
    ll ans = 0;
    for (int j=0; j < m; j++) {
        if (v[i][j] == sp[cc]) {
            if (cc == 4) {
                ans += 5;
            }
            cc = (cc+1) % 5;
        } else {
            if (sps.find(v[i][j]) != sps.end()) {
                ans--;
            }
        }
    }
    return {ans, cc};
}

pair<ll,bool> cal(int i, int pr) {
    if (i >= n) {
        return {0,false};
    }
    if (do1[i][pr] == tt1) {
        return dp[i][pr];
    }
    auto p = inteScore(i, pr);
    auto op1 = cal(i+1, p.second);
    auto op2 = cal(i+1, pr);
    ll ans1 = p.first + op1.first + (op1.second ? 0 : -p.second);
    ll ans2 = cal(i+1, pr).first;
    ll tm = max(ans1, ans2);
    do1[i][pr] = tt1;
    bool picked = (ans1 > ans2) || op2.second;
    return dp[i][pr] = {tm, picked};
}
void solve(int tt) {
    cin >> n >> m;
    tt1 = tt+1;
    v.clear();
    for (int i=0; i < n; i++) {
        string s;
        cin >> s;
        v.push_back(s);
    }
    cout<<cal(0, 0).first<<"\n";
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    ll test_cases = 1;
    cin >> test_cases;
    for(int tc=0;tc<test_cases;tc++){
        solve(tc);
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
s = 'narek'

def fun(i, j):
    if i < 0:
        return -(4-j)
    # if dp[i][j] != -1:
    #     return dp[i][j]
    m1 = fun(i - 1, j)
    x = 0
    for k in range(m - 1, -1, -1):
        if a[i][k] in s:
            if a[i][k] == s[j]:
                if j == 0:
                    x += 5
                    j = 4
                else:
                    j -= 1
            else:
                x -= 1
    m2 = fun(i - 1, j) + x
    # dp[i][j] = max(m1, m2)
    # print(i, m1, m2, x)
    return max(m1, m2)

for _ in range(int(input())):
    n, m = map(int, input().split())
    a = [input() for __ in range(n)]
    # dp = [[-1] * 5 for _ in range(n)]
    print(fun(n - 1, 4))

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5 5
narek
narek
narek
narek
narek
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
import java.io.*;
import java.util.*;

public class Main {

    static FastScanner fs = new FastScanner();
    static final Random random = new Random();

    static int inf = 3000000 ;
    static int N = 2000001 ;
    static int mod = 1000000007 ;
    static int chars = 26 ;

    static String narek = "narek" ;

    static int on ( String s  , int l , int r ) {
        if ( r < l ) return 0 ;
        int ans = 0 ;
        for ( int i = l ; i <= r ; i ++ ) {
            for ( int j = 0 ; j < narek.length() ; j ++ ) {
                if ( s.charAt(i) == narek.charAt(j) ) {
                    ans ++ ;
                    break ;
                }
            }
        }
        return ans ;
    }

    static int rec ( int i , int j , int n , int m , String[] str , String all ) { // Approach Test
        if ( i == n * m ) {
            return j == 0 ? 0 : -inf ;
        }
        int grp = i / m ;
        int ans = - inf ;
        if ( all.charAt(i) == narek.charAt(j) ) {
            for ( int k = i + 1 ; k < ( grp + 1 ) * m ; k ++ ) { // same segment
                int res = rec ( k , ( j + 1 ) % narek.length() , n , m , str , all ) ;
                if ( res != - inf  ) {
                    ans = Math.max ( ans , 1 - on ( all ,  i + 1 , k - 1 ) + res ) ;
                }
            }

            for ( int k = ( grp + 1 ) * m ; k <= n * m ; k ++ ) { // segemnt change
                int res = rec ( k , ( j + 1 ) % narek.length() , n , m , str , all ) ;
                if ( res != -inf  ) {
                    int nxtGrp = ( k / m ) ;
                    ans = Math.max ( ans , 1 - on ( all , i + 1 , ( ( 1 + grp ) * m ) - 1 ) - on ( all , ( nxtGrp ) * m , k - 1 ) + res ) ;
                }
            }
        }
        return ans ;
    }

    static void Test_Case( int t ) {
        int n = fs.nextInt() , m = fs.nextInt() ;
        String[] str = new String[n] ;
        String all = "" ;
        for ( int i = 0 ; i < n ; i ++ ) {
            str[i] = fs.next() ;
            all = all + str[i] ;
        }
        int ans = 0 ;
        for ( int i = 0 ; i < n * m ; i ++ ) {
            int res = rec ( i , 0 , n , m , str , all ) ;
            if ( res != - inf ) {
                int grp = i / m ;
                int here = res - on ( all , ( grp ) * m , i - 1 ) ;
                ans = Math.max(ans, here ) ;
            }
        }
        System.out.println( ans );
    }

    public static void main(String[] useCppForCp ) {
        // precompFacts() ;
        int t = 1 ;
        t = fs.nextInt();
        while ( t > 0  ) {
            Test_Case( t ) ;
            t -- ;
        }
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }

    static long add(long a, long b) {
		return (a + b) % mod;
	}

	static long sub(long a, long b) {
		return ((a - b) % mod + mod) % mod;
	}

	static long mul(long a, long b) {
		return ((a % mod ) * (b % mod )) % mod;
	}

    static int mul(int a, int b) {
		return (a % mod* b % mod) % mod;
	}

	static long exp(long base, long exp) {
		if (exp == 0)
			return 1;
		long half = exp(base, exp / 2);
		if (exp % 2 == 0)
			return mul(half, half);
		return mul(half, mul(half, base));
	}

	static long[] factorials = new long[N];
	static long[] invFactorials = new long[N];
    static long[] modInv = new long[N] ;

	static void precompFacts() {
		factorials[0] = invFactorials[0] = 1;
		for (int i = 1; i < factorials.length; i++)
			factorials[i] = mul(factorials[i - 1], i);
		invFactorials[factorials.length - 1] = exp(factorials[factorials.length - 1], mod - 2);
		for (int i = invFactorials.length - 2; i >= 0; i--)
			invFactorials[i] = mul(invFactorials[i + 1], i + 1);
        for ( int i = 0 ; i < modInv.length ; i ++ ) {
            modInv[i] = modInverse(i, mod) ;
        }
	}

	static long nCk(int n, int k) {
		return mul(factorials[n], mul(invFactorials[k], invFactorials[n - k]));
	}

    public static long modDivide(long a, int b) {
        long b_inv = modInv[b];
        return (a * b_inv) % mod;
    }

    public static long modInverse(long b, long mod) {
        return power(b, mod - 2, mod);
    }

    public static long power(long base, long exponent, long mod) {
        long result = 1;
        base = base % mod;

        while (exponent > 0) {
            if ((exponent & 1) == 1) {
                result = (result * base) % mod;
            }

            exponent = exponent >> 1; // Divide exponent by 2
            base = (base * base) % mod; // Change base to base^2
        }
        return result;
    }

}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that can highlight issues in the algorithm's logic. Here are the key points to consider:

1. **Edge Cases:**
   - Strings that do not contain all the letters of "narek" but contain some of them.
   - Strings that contain all the letters of "narek" multiple times.
   - Strings that contain letters in a non-consecutive order.
   - Strings that contain a mix of letters, some of which are part of "narek" and some of which are not.

2. **Test Input Design:**
   - We will create a test case where Narek can potentially find multiple "narek" sequences, but the way the code handles the segments and the transition between them might be flawed.
   - We will also include a string that contains all the letters of "narek" but in a different order to test the code's ability to recognize and handle such cases.

Here is a small-scale test input that should expose potential issues in the buggy code:

```plaintext
4
5 5
narek
narek
narek
narek
narek
5 5
narek
knear
arenk
enark
rnkea
3 5
narek
nare
nark
1 5
narek
```

**Explanation:**
1. **First Test Case:**
   - All strings are "narek". This should be an easy case where Narek can find multiple "narek" sequences.
   - Expected output: 25 (5 * 5 - 0)

2. **Second Test Case:**
   - Each string contains the letters of "narek" but in different orders. This tests the code's ability to handle non-consecutive sequences.
   - Expected output: 25 (5 * 5 - 0)

3. **Third Test Case:**
   - The first string is "narek", the second string is "nare" (missing 'k'), and the third string is "nark" (missing 'e').
   - Expected output: 10 (2 * 5 - 0)

4. **Fourth Test Case:**
   - A single string "narek". This is a simple case to ensure the code handles a single string correctly.
   - Expected output: 5 (1 * 5 - 0)

This test input should help identify any issues in the buggy code's logic, especially in how it handles the transition between segments and the recognition of non-consecutive sequences.

4
5 5
narek
narek
narek
narek
narek
5 5
narek
knear
arenk
enark
rnkea
3 5
narek
nare
nark
1 5
narek


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Lazy Narek
time_limit_ms: 2000
memory_limit_mb: 256
Description: Narek is too lazy to create the third problem of this contest. His friend Artur suggests that he should use ChatGPT. ChatGPT creates $n$ problems, each consisting of $m$ letters, so Narek has $n$ strings. To make the problem harder, he combines the problems by selecting some of the $n$ strings possibly none and concatenating them without altering their order. His chance of solving the problem is defined as $score_n - score_c$, where $score_n$ is Narek's score and $score_c$ is ChatGPT's score.

Narek calculates $score_n$ by examining the selected string (he moves from left to right). He initially searches for the letter $\texttt{"n"}$, followed by $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, and $\texttt{"k"}$. Upon finding all occurrences of these letters, he increments $score_n$ by $5$ and resumes searching for $\texttt{"n"}$ again (he doesn't go back, and he just continues from where he left off).

After Narek finishes, ChatGPT scans through the array and increments $score_c$ by $1$ for each letter $\texttt{"n"}$, $\texttt{"a"}$, $\texttt{"r"}$, $\texttt{"e"}$, or $\texttt{"k"}$ that Narek fails to utilize (note that if Narek fails to complete the last occurrence by finding all of the $5$ letters, then all of the letters he used are counted in ChatGPT's score $score_c$, and Narek doesn't get any points if he doesn't finish finding all the 5 letters).

Narek aims to maximize the value of $score_n - score_c$ by selecting the most optimal subset of the initial strings.

Input: In the first line of the input, you're given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you're given two integers $n, m$ ($1 \le n, m \le 10^3$), the number of strings and the length of each string.

In the next $n$ lines, you're given $n$ strings, each having a length of $m$. The strings only contain lowercase letters of the English alphabet.

The sum of values of $n \cdot m$ over all test cases does not exceed $10^6$.

Output: For each test case, output a single integer: the maximal possible value of $score_n - score_c$.

Examples:
input:
4
5 2
nn
aa
rr
ee
kk
1 5
narek
1 4
nare
5 7
nrrarek
nrnekan
uuuuuuu
ppppppp
nkarekz
output:
0
5
0
7

Note: In the first test case, one of the optimal answers is when Narek doesn't choose any of the strings, so the answer is $0$. He can alternatively choose all the strings. In this case, the full string becomes "nnaarreekk". Narek can choose the first appearances of all letters and add $5$ to the score. His opponent will add $1$ for all second appearances, which will be $5$ in total. So the answer will be $5 - 5 = 0$.

In the third test case, the only optimal answer is when Narek doesn't choose the string. Note that if he were to choose the string, he wouldn't be able to find the last letter "k", so his score would stay at $0$ instead of becoming $5$. Then ChatGPT would add $4$ for all of the $4$ letters, and the answer would become $0 - 4 = -4$.

In the last test case, Narek needs to choose the first and the last strings. After putting these two next to each other, he gets "${\color{red}{n}}rr{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{red}{n}}k{\color{red}{a}}{\color{red}{r}}{\color{red}{e}}{\color{red}{k}}{\color{blue}{z}}$". Narek can choose the letters marked with red and add $10$ to his score. Since the black colored letters Narek left behind are eligible for the opponent to claim (they are used in the word "narek"), the opponent adds all other letters to the score and gets a score of $3$. Therefore, the answer is $10 - 3 = 7$.

Buggy Code:
import java.util.*;

public class NarekVsChatGPT {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int testCases = scanner.nextInt();

        while (testCases-- > 0) {
            int stringCount = scanner.nextInt();
            int stringLength = scanner.nextInt();

            String[] words = new String[stringCount];
            for (int i = 0; i < stringCount; i++) {
                words[i] = scanner.next();
            }

            int n = words.length;
            int maxScore = Integer.MIN_VALUE;
            int[] scores = new int[1 << n];

            for (int mask = 1; mask < (1 << n); mask++) {
                StringBuilder combined = new StringBuilder();

                for (int i = 0; i < n; i++) {
                    if ((mask & (1 << i)) != 0) {
                        combined.append(words[i]);
                    }
                }

                int narekPoints = 0, chatGPTPoints = 0;
                int nCount = 0, aCount = 0, rCount = 0, eCount = 0, kCount = 0;
                for (char c : combined.toString().toCharArray()) {
                    if (c == 'n' && nCount == 0) nCount++;
                    else if (c == 'a' && nCount == 1 && aCount == 0) aCount++;
                    else if (c == 'r' && aCount == 1 && rCount == 0) rCount++;
                    else if (c == 'e' && rCount == 1 && eCount == 0) eCount++;
                    else if (c == 'k' && eCount == 1 && kCount == 0) {
                        kCount++;
                        narekPoints += 5;
                        nCount = aCount = rCount = eCount = kCount = 0;
                    } else if ("narek".indexOf(c) != -1) {
                        chatGPTPoints++;
                    }
                }

                chatGPTPoints += nCount + aCount + rCount + eCount + kCount;

                scores[mask] = narekPoints - chatGPTPoints;
            }

            for (int score : scores) {
                maxScore = Math.max(maxScore, score);
            }

            System.out.println(maxScore);
        }

        scanner.close();
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5 5
narek
narek
narek
narek
narek
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimize!
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $$$a$$$ and $$$b$$$ ($$$a \leq b$$$). Over all possible integer values of $$$c$$$ ($$$a \leq c \leq b$$$), find the minimum value of $$$(c - a) + (b - c)$$$.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 55$$$) — the number of test cases.

Each test case contains two integers $$$a$$$ and $$$b$$$ ($$$1 \leq a \leq b \leq 10$$$).

Output: For each test case, output the minimum possible value of $$$(c - a) + (b - c)$$$ on a new line.

Examples:
input:
3
1 2
3 10
5 5
output:
1
7
0

Note: In the first test case, you can choose $$$c = 1$$$ and obtain an answer of $$$(1 - 1) + (2 - 1) = 1$$$. It can be shown this is the minimum value possible.

In the second test case, you can choose $$$c = 6$$$ and obtain an answer of $$$(6 - 3) + (10 - 6) = 7$$$. It can be shown this is the minimum value possible.

Buggy Code:
#include<stdio.h>

int main()
{
    int t,a,b;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&a,&b);

        for(int c=0;c<b;c++)
        {
            int v=(c-a)+(b-c);
            if(a==b)
            {
                printf("0\n");
                break;
            }
            if(a<=v && b>=v)
            {
                printf("%d\n",v);
                break;
            }
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
5
1 1
1 2
2 2
1 10
5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimize!
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two integers $$$a$$$ and $$$b$$$ ($$$a \leq b$$$). Over all possible integer values of $$$c$$$ ($$$a \leq c \leq b$$$), find the minimum value of $$$(c - a) + (b - c)$$$.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 55$$$) — the number of test cases.

Each test case contains two integers $$$a$$$ and $$$b$$$ ($$$1 \leq a \leq b \leq 10$$$).

Output: For each test case, output the minimum possible value of $$$(c - a) + (b - c)$$$ on a new line.

Examples:
input:
3
1 2
3 10
5 5
output:
1
7
0

Note: In the first test case, you can choose $$$c = 1$$$ and obtain an answer of $$$(1 - 1) + (2 - 1) = 1$$$. It can be shown this is the minimum value possible.

In the second test case, you can choose $$$c = 6$$$ and obtain an answer of $$$(6 - 3) + (10 - 6) = 7$$$. It can be shown this is the minimum value possible.

Buggy Code:
#include<stdio.h>
int main(void){
	int c[55];
	int a,b,n,sum=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d %d",&a,&b);
		for(int j=a;j<=b;j++){
			int sum1=(j-a)+(b-j);
			if(sum1>sum){
				sum=sum1;
				c[i]=sum;
			}
		}
	}
	for(int d=1;d<=n;d++){
		printf("%d\n",c[d]);
	}
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
5
1 1
1 2
2 2
1 10
10 10
```