**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
import java.util.Scanner;

public class A13 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();
        int f = 1;

        while(t-->0){

            int n = sc.nextInt(),b=sc.nextInt(),a=sc.nextInt();
            if(f == 134)System.out.println(n+" "+a+" "+b);
            f++;

            String str = sc.next();

            int s1=0,s2=0;
            boolean ok = false;

            for(int i = 0;i<n;i++){
                char ch = str.charAt(i);
                if(ch == 'N')s1++;
                else if(ch == 'S')s1--;
                else if(ch == 'E')s2++;
                else s2--;
                if(a == s1 && b == s2){System.out.println("YES");ok = true;break;}
            }
            if(ok)continue;
            int curr1 = s1,curr2 = s2;
            int k = 0;

            while(k<30){
                for(int i = 0;i<n;i++){
                    char ch = str.charAt(i);
                    if(ch == 'N')curr1++;
                    else if(ch == 'S')curr1--;
                    else if(ch == 'E')curr2++;
                    else curr2--;
                    if(a == curr1 && b == curr2){System.out.println("YES");ok = true;break;}
                }
                if(ok)break;
                curr1 += s1;curr2 += s2;
                k++;
            }
            if(!ok)System.out.println("NO");
        }
        sc.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
3
1 1 1
N
1 1 1
E
1 1 1
S
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
import java.util.Scanner;

public class AliceAndRedQueen {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();  // number of test cases

        while (t-- > 0) {
            int n = scanner.nextInt();  // length of the move sequence
            int a = scanner.nextInt();  // Red Queen's x-coordinate
            int b = scanner.nextInt();  // Red Queen's y-coordinate
            String s = scanner.next();  // move sequence

            // Initialize Alice's position
            int x = 0, y = 0;

            // Calculate the net displacement in one cycle of s
            int dx = 0, dy = 0;
            for (char move : s.toCharArray()) {
                switch (move) {
                    case 'N': dy++; break;
                    case 'E': dx++; break;
                    case 'S': dy--; break;
                    case 'W': dx--; break;
                }
            }

            // Track Alice's visited positions in one cycle
            boolean met = false;
            int cycleLength = s.length();
            for (int i = 0; i < cycleLength; i++) {
                // Check if current position matches Red Queen's position
                if (x == a && y == b) {
                    met = true;
                    break;
                }

                // Move according to the ith character in s
                char move = s.charAt(i);
                switch (move) {
                    case 'N': y++; break;
                    case 'E': x++; break;
                    case 'S': y--; break;
                    case 'W': x--; break;
                }
            }

            // If met within one cycle, print "YES" and continue
            if (met) {
                System.out.println("YES");
                continue;
            }

            // If net displacement is zero, Alice moves in a loop that doesn't go anywhere
            if (dx == 0 && dy == 0) {
                System.out.println("NO");
                continue;
            }

            // Simulate additional cycles
            int currentX = x, currentY = y;
            for (int cycle = 1; cycle <= 500; cycle++) {
                currentX += dx;
                currentY += dy;

                // Check if Red Queen's position is reached after applying the net displacement
                if (currentX == a && currentY == b) {
                    met = true;
                    break;
                }
            }

            System.out.println(met ? "YES" : "NO");
        }

        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
5
1 1 1
N
2 2 2
NE
4 5 5
NEWS
3 3 3
NEE
6 6 6
NNEESW
```

5
1 1 1
N
2 2 2
NE
4 5 5
NEWS
3 3 3
NEE
6 6 6
NNEESW


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
      int a,b,n;
      cin>>n>>a>>b;
      string s;
      cin>>s;

      int x=0,y=0;
      int f=0;
      while(true){
      for(int i=0;i<n;i++){
        if(a==x&&b==y){
           f=1;
           break;
        }
         if(s[i]=='N') y++;
         if(s[i]=='S') y--;
         if(s[i]=='E') x++;
         if(s[i]=='W') x--;
      }
      if(x>10||y>10) break;
      if(f==1){
        break;
      }
      if(x<=0||y<=0){
         break;
      }
      }
      if(f==1){
        cout<<"YES"<<endl;
      }
      else{
        cout<<"NO"<<endl;
      }
    }
    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
N
1 1 1
S
1 1 1
E
1 1 1
W
10 10 10
NNEESWNEWS
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
#include<iostream>
#include <vector>
#include<string>

using namespace std;
int main() {
    int a;
    cin>>a;
    while(a--) {
        int n,m,o;
        cin>>n>>m>>o;
        string t;
        cin>>t;
        vector<pair<int,int>> v(n);
        pair<int,int> p = {0,0};
        for(int i=0;i<n;i++) {
            if(t[i] == 'E') {
                p.first++;
            }else if(t[i] == 'W') {
                p.first--;
            }else if(t[i] == 'N') {
                p.second++;
            }else {
                p.second--;
            }
            v[i] = p;
        }

        bool flag = false;
        if(p.first==0 && p.second==0) {
            if(m ==0 && o ==0) {
                flag = true;
            }else {
                for(int i = 0;i<n-1;i++) {
                    if(m == v[i].first && o == v[i].second) {
                        flag = true;
                        break;
                    }
                }
            }
        }else if(p.first ==0) {
            if(m == 0&& o%p.second ==0) {
                flag = true;

            }else {
                for(int i = 0;i<n-1;i++) {
                    if(m == v[i].first && o%v[i].second ==0) {
                        flag = true;
                        break;
                    }
                }
            }
        }else if(p.second ==0) {
                if(m%p.first == 0&&o == 0) {
                    flag = true;
                }else {
                    for(int i = 0;i<n-1;i++) {
                        if(m%v[i].first == 0 && o == v[i].second) {
                            flag = true;
                            break;
                        }

                    }
                }
            }else {
                if((m%p.first ==0)&&(o%p.second==0)&&(o/p.second==m/p.first)) {
                    flag = true;
                }else {
                    for(int i = 0;i<n-1;i++) {
                        if(((m-v[i].first)%p.first ==0)&&((o-v[i].second)%p.second==0)&&((m-v[i].first)/p.first==(o-v[i].second)/p.second)) {
                            flag = true;
                            break;
                        }
                    }
                }
            }
        if(flag){
            cout<<"YES"<<endl;
        }else {
            cout<<"NO"<<endl;
        }
        }
    }

**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
4 1 1
NEWS
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
t = int(input())

while t>0:
    n,a,b  = map(int,input().split())
    st  = input()
    x,y=0,0
    ok = False
    for ch in st:
        if ch=='N':
            y+=1
        if ch=='S':
            y-=1
        if ch=='E':
            x+=1
        if ch=='W':
            x-=1
        if x==a and y==b:
            ok = True
            break

    if a==b:
        if x==y and (x!=0):
            print("YES")
        else:
            print("NO")
    elif a==x and y==b:
        print("YES")
    elif a>b:
        if x>y and (x<a):
            if (a-x)%(b-y)==0:
                print("YES")
            else:
                print("NO")
        else:
            print("NO")
    else:
        if y>x and y<b:
            if (b-y)%(a-x)==0:
                print("YES")
            else:
                print("NO")
        else:
            print("NO")
    t-=1

**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
6
1 1 1
N
1 1 1
E
1 1 1
S
1 1 1
W
2 2 2
NE
3 3 3
NNE
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
def solve(n,a,b):
    string=input()
    x=0
    y=0
    for char in string:
        if char =="N":
            y+=1
        elif char =="S":
            y-=1
        elif char =="E":
            x+=1
        else:
            x-=1
    if x<=0 and y<=0:
        return "NO"
    while a>x and b>y:
        a-=x
        b-=y
    for char in string:
        if char =="N":
            b-=1
        elif char =="S":
            b+=1
        elif char =="E":
            a-=1
        else:
            a+=1
        if a==0 and b==0:
            return "YES"
    return "NO"

t = int(input())
for i in range(t):
    n , a , b = map(int,input().split())
    answer = solve(n, a ,b)
    print(answer)

**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
1 1 1
N
2 1 1
NE
3 1 1
NNE
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
import java.util.Scanner;

public class A {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int test = sc.nextInt();

        StringBuilder output = new StringBuilder("");
        while(test-- > 0) {
            int n = sc.nextInt();
            int a = sc.nextInt();
            int b = sc.nextInt();
            sc.nextLine();

            int x = 0;
            int y = 0;
            String str = sc.nextLine();
            boolean flag = true;

            if (a == 0 && b == 0) {
                output.append("YES").append("\n");
                continue;
            }

                for (int i = 0; i < n; i++) {
                    char c = str.charAt(i);
                    if (c == 'W') {
                        x--;
                    } else if (c == 'E') {
                        x++;
                    } else if (c == 'N') {
                        y++;
                    } else {
                        y--;
                    }

                    if (x == a && y == b) {
                        output.append("YES").append("\n");
                        flag = false;
                        break;
                    }
                }

            if (!flag) {
                continue;
            }

            if (flag) {
                if (x == 0 && y == 0) {
                    output.append("NO").append("\n");
                    continue;
                }

                int temp = a > b ? b : a;
                for(int k = b; k >= 2; k--) {
                    if (a % k == 0 && b % k == 0) {
                        a /= k;
                        b /= k;
                    }
                }
                if (a == x && b == y) {
                    output.append("YES").append("\n");
                    continue;
                }

                if (x < 0 && y >= 0) {
                    output.append("YES").append("\n");
                    continue;
                }

                if (y < 0 && x >= 0) {
                    output.append("NO").append("\n");
                    continue;
                }
            }

            output.append("NO").append("\n");

        }

        System.out.println(output);

    }
}

**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
1 1 1
N
1 1 1
E
1 1 1
S
1 1 1
W
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
import java.util.Scanner;

public class alice_adventures_in_chess {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int a = sc.nextInt();
            int b = sc.nextInt();

            int x = 0, y = 0;
            sc.nextLine();
            String s = sc.nextLine();

            for (int i = 0; i < n; i++) {
                char c = s.charAt(i);
                if (c == 'N') {
                    y++;
                }
                if (c == 'E') {
                    x++;
                }
                if (c == 'W') {
                    x--;
                }
                if (c == 'S') {
                    y--;
                }
            }

            int final_x = x, final_y = y;
            x = 0;
            y = 0;

            boolean found = false;
            for (int i = 0; i < n; i++) {
                if (final_x != 0 && (a - x) % final_x == 0 && (a - x) / final_x == (b - y) / final_y) {
                    if (final_y != 0 && (b - y) % final_y == 0) {
                        System.out.println("yes");
                        found = true;
                        break;
                    }
                }
                char c = s.charAt(i);
                if (c == 'N') {
                    y++;
                }
                if (c == 'E') {
                    x++;
                }
                if (c == 'W') {
                    x--;
                }
                if (c == 'S') {
                    y--;
                }
            }

            if (!found) {
                System.out.println("no");
            }
        }

        sc.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
3
4 1 1
NEWS
4 2 2
NEES
4 3 3
NNEE
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long int
using namespace std;
int main() {
    int t;
    cin>>t;
    while(t--){
        int n,a,b;
        string s1;
        cin>>n>>a>>b;
                cin>>s1;
        int fst=abs(a);
        int snd=abs(b);
        int x=0,y=0;
        int flag=0;
        for(int i=0;i<n+100;i++){
            char c;
            c=s1[i%n];
            if(c=='N'){
                ++y;
            }else{
              if(c=='E')++x;
              if(c=='S')--y;
              if(c=='W')--x;
            }
           // cout<<x<<" "<<y<<endl;
            if(x==fst && y==snd)flag=1;
        }
        if(flag)cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }

}

**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
4 0 0
NEWS
```

### Explanation:
- **Test Case:**
  - **n = 4**: The length of the string `s1`.
  - **a = 0**: The x-coordinate of the Red Queen.
  - **b = 0**: The y-coordinate of the Red Queen.
  - **s1 = "NEWS"**: The sequence of moves Alice will follow.

- **Expected Output:**
  - **"YES"**: Alice should meet the Red Queen at (0, 0).

### Reasoning:
1. **Initial Position:**
   - Alice starts at (0, 0).

2. **Sequence of Moves:**
   - **N**: Move to (0, 1).
   - **E**: Move to (1, 1).
   - **W**: Move to (0, 1).
   - **S**: Move to (0, 0).

3. **Result:**
   - After one complete cycle of the sequence "NEWS", Alice returns to the starting position (0, 0), which is the same as the Red Queen's position.
   - Therefore, Alice will meet the Red Queen.

### Bug Exposure:
- The buggy code does not correctly handle the case where the Red Queen is at the starting position (0, 0). The code initializes `fst` and `snd` as the absolute values of `a` and `b`, which is unnecessary and can lead to incorrect comparisons. This test case exposes the issue by showing that the code should recognize Alice meeting the Red Queen at the starting position.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, a, b;
        cin >> n >> a >> b;
        char arr[n];
        int x = 0;
        int y = 0;
        for (int i = 0; i < n; i++)
        {
            cin >> arr[i];
            if (arr[i] == 'N')
            {
                y++;
            }
            else if (arr[i] == 'S')
            {
                y--;
            }
            else if (arr[i] == 'E')
            {
                x++;
            }
            else if (arr[i] == 'W')
            {
                x--;
            }
        }
        int x_div = x ? a / x : 0;
        int y_div = y ? b / y : 0;
        int x_rem = x ? a % x : 0;
        int y_rem = x ? a % x : 0;
        int mul = max(0, min(x_div - 1, y_div - 1));
        x = x * mul;
        y = y * mul;
        int isfalse = 0;
        if (x == a && y == b)
        {
            isfalse = 1;
        }
        else
        {
            for (int i = 0; i < n; i++)
            {
                if (arr[i] == 'N')
                {
                    y++;
                }
                else if (arr[i] == 'S')
                {
                    y--;
                }
                else if (arr[i] == 'E')
                {
                    x++;
                }
                else if (arr[i] == 'W')
                {
                    x--;
                }
                if (x == a && y == b)
                {
                    isfalse = 1;
                    break;
                }
            }
            for (int i = 0; i < n; i++)
            {
                if (arr[i] == 'N')
                {
                    y++;
                }
                else if (arr[i] == 'S')
                {
                    y--;
                }
                else if (arr[i] == 'E')
                {
                    x++;
                }
                else if (arr[i] == 'W')
                {
                    x--;
                }
                if (x == a && y == b)
                {
                    isfalse = 1;
                    break;
                }
            }
            for (int i = 0; i < n; i++)
            {
                if (arr[i] == 'N')
                {
                    y++;
                }
                else if (arr[i] == 'S')
                {
                    y--;
                }
                else if (arr[i] == 'E')
                {
                    x++;
                }
                else if (arr[i] == 'W')
                {
                    x--;
                }
                if (x == a && y == b)
                {
                    isfalse = 1;
                    break;
                }
            }
            for (int i = 0; i < n; i++)
            {
                if (arr[i] == 'N')
                {
                    y++;
                }
                else if (arr[i] == 'S')
                {
                    y--;
                }
                else if (arr[i] == 'E')
                {
                    x++;
                }
                else if (arr[i] == 'W')
                {
                    x--;
                }
                if (x == a && y == b)
                {
                    isfalse = 1;
                    break;
                }
            }
        }
        if (isfalse == 1)
        {
            cout << "YES" << endl;
        }
        else
        {
            cout << "NO" << endl;
        }
    }
}

**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
1 1 1
N
2 2 2
NE
4 5 5
NEWS
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
import sys

input = lambda: sys.stdin.readline().strip()

T = int(input())
for _ in range(T):
    n, a, b = map(int, input().split())
    s = input()
    ans = 'NO'
    x = 0
    y = 0
    for i in range(11):
        for j in s:
            if j == 'N':
                y += 1
            elif j == 'S':
                y -= 1
            elif j == 'E':
                x += 1
            else:
                x -= 1
            if x == a and y == b:
                ans = 'YES'
    print(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
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
```

6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
for _ in range(int(input())):
    n, a, b = [int(x)for x in input().split()]
    s = input()

    s = s * 10

    aliceX, aliceY = 0, 0
    ans = False

    for i in s:
        if i == "N":
            aliceY += 1
        elif i == "E":
            aliceX += 1
        elif i == "S":
            aliceY -= 1
        else:
            aliceX -= 1

        if (aliceX == a) and (aliceY == b):
            ans = True
            break

    if ans:
        print("YES")
    else:
        print("NO")

**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
1 1 1
N
1 1 1
E
1 1 1
W
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
import java.util.*;

public class Contest {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int testcases = sc.nextInt();

        for (int z = 0; z < testcases; z++) {

            sc.nextInt();

            int a = sc.nextInt();
            int b = sc.nextInt();

            String s = sc.next();

            int temp = 100;
            boolean found = false;

            while (temp-- > 0) {
                for (char c : s.toCharArray()) {
                    switch (c) {
                        case 'N':
                            b--;
                            break;

                        case 'S':
                            b++;
                            break;

                        case 'E':
                            a--;
                            break;

                        case 'W':
                            a++;
                            break;

                        default:
                            break;
                    }

                    if (a == 0 && b == 0) {
                        System.out.println("YES");
                        found = true;
                    }
                }

                if (found == true || (a < 0 || b < 0)) {
                    break;
                }
            }

            if (found == false) {System.out.println("NO");}
        }

        sc.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
3
1 1 1
N
1 1 1
E
1 1 1
S
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
import java.util.Scanner;

public class Minnu {

public static void main(String[] args) {
 Scanner s=new Scanner(System.in);
 int t=s.nextInt();
   l: for (int i = 0; i < t; i++) {
        int n=s.nextInt();
        int a=s.nextInt();
        int b=s.nextInt();
        String ans=s.next();
        int c=0;
        int d=0;
        int j=0;
        int x=a;
        int y=b;
        int k=100;
        while(k!=0){
            char ch=ans.charAt(j%n);
            if(ch=='N'){
                d++;
            }else if(ch=='E'){
                c++;
            }else if(ch=='S'){
                d--;
            }else{
                c--;
            }
            j++;
            k--;
            if(a==c && b==d){
                System.out.println("yes");
                continue l;
            }
        }
        System.out.println("no");
    }
}
    }

**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 expose the bug in the given code, we need to focus on scenarios where the code might fail. The primary issues in the code are:

1. **Infinite Loop or Long Running Time**: The code uses a fixed number of iterations (100) to check if Alice meets the Red Queen. This might not be sufficient in some cases, and it might also run longer than necessary in other cases.
2. **Incorrect Position Handling**: The code does not correctly handle the periodic nature of Alice's movements. It should consider the net effect of the movements over the length of the string and then check if this net effect can reach the Red Queen's position.

To create a test case that exposes these issues, we can:

1. Use a small string that has a net effect that does not immediately reach the Red Queen but could over multiple cycles.
2. Use a position for the Red Queen that is reachable only after several cycles of the movement pattern.

Here's a test input that should expose the bug:

```plaintext
3
4 4 4
NEES
4 5 5
NEWS
6 3 3
NNEESS
```

Explanation:
1. **Test Case 1**: Alice's movement pattern is "NEES". This pattern results in a net movement of (0, 0) after one cycle. Therefore, Alice will never reach (4, 4), but the code might incorrectly output "yes" if it runs for too few iterations.
2. **Test Case 2**: Alice's movement pattern is "NEWS". This pattern results in a net movement of (0, 0) after one cycle. Therefore, Alice will never reach (5, 5), but the code might incorrectly output "yes" if it runs for too few iterations.
3. **Test Case 3**: Alice's movement pattern is "NNEESS". This pattern results in a net movement of (0, 2) after one cycle. Therefore, Alice will reach (3, 3) after 1.5 cycles (3 moves in the north direction and 3 moves in the east direction). The code should output "yes" but might fail if it runs for too few iterations.

This test input should help identify the issues with the code's fixed iteration count and incorrect handling of periodic movements.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
#include <iostream>
#include <algorithm>
using namespace std;

void solve() {
    int n;
    cin >> n;
    int arr[200005];
    for (int i = 1; i <= n; i++) cin >> arr[i];
    sort(arr + 1, arr + n + 1);
    int a = -1e9, b = -1e9, c = -1e9, d = -1e9, c1 = 0, c2 = 0;
    for (int i = 1; i < n; i++) {
        if (arr[i] == arr[i + 1]) {
            a = arr[i];
            c1 = i + 2;
            break;
        }
    }
    for (int i = c1; i < n; i++) {
        if (arr[i] == arr[i + 1]) {
            b = arr[i];
            c2 = i + 2;
            break;
        }
    }
    for (int i = n; i > c2; i--) {
        if (arr[i] == arr[i - 1]) {
            d = arr[i];
            c1 = i - 2;
            break;
        }
    }
    for (int i = c1; i > c2; i--) {
        if (arr[i] == arr[i - 1]) {
            c = arr[i];
            break;
        }
    }
    if (a <= b && b <= c && c <= d) {
        cout << "YES" << endl;
        cout << a << " " << b << " " << a << " " << d << " " << c << " " << b << " " << c << " " << d << endl;
    } else cout << "NO" << endl;
}

int main() {
    int t;
    cin >> t;
    while (t--) solve();
    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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**:
   - Cases where there are not enough repeated elements to form a rectangle.
   - Cases where the repeated elements are not in the correct order to form a valid rectangle.
   - Cases where the elements are very close to each other to test the precision of the algorithm.

2. **Input Constraints**:
   - The input should be small but valid, with \( t \) (number of test cases) and \( n \) (number of elements) within the specified limits.
   - The elements \( a_i \) should be within the range \([-10^9, 10^9]\).

3. **Buggy Code Analysis**:
   - The code tries to find pairs of equal elements to form the coordinates of the rectangle.
   - It assumes that the pairs are in a specific order to form a valid rectangle, which might not always be the case.

Given these points, let's create a test input that exposes the bug:

```plaintext
3
8
1 1 2 2 3 3 4 4
8
1 2 3 4 5 6 7 8
10
1 1 2 2 3 3 4 4 5 5
```

### Explanation:
1. **First Test Case**:
   - The list is `[1, 1, 2, 2, 3, 3, 4, 4]`.
   - There are repeated elements, and a valid rectangle can be formed with coordinates (1, 2, 1, 4, 3, 2, 3, 4).
   - Expected output: `YES` followed by the coordinates.

2. **Second Test Case**:
   - The list is `[1, 2, 3, 4, 5, 6, 7, 8]`.
   - There are no repeated elements, so it is impossible to form a rectangle.
   - Expected output: `NO`.

3. **Third Test Case**:
   - The list is `[1, 1, 2, 2, 3, 3, 4, 4, 5, 5]`.
   - There are repeated elements, and a valid rectangle can be formed with coordinates (1, 2, 1, 4, 3, 2, 3, 4).
   - Expected output: `YES` followed by the coordinates.

This test input should effectively expose the bug in the buggy code by including cases where the repeated elements are not in the correct order to form a valid rectangle.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
#include<iostream>
#include<map>
using namespace std;
const int N=2e5+10;
int t,n,a,p[4],cnt;
map<int,int>mp;
int main(){
	cin>>t;
	while(t--){
		mp.clear();
		cin>>n;
		for(int i=1;i<=n;i++){
			cin>>a;
			mp[a]++;
		}
		cnt=0;
		for(map<int,int>::iterator it=mp.begin();it!=mp.end();it++){
			if((it->second)>=2){
				p[cnt++]=it->first;
				mp[it->first]-=2;
			}
			if(cnt>1)break;
		}
		for(map<int,int>::reverse_iterator it=mp.rbegin();it!=mp.rend();it++){
			if((it->second)>=2){
				p[cnt++]=it->first;
				mp[it->first]-=2;
			}
			if(cnt>3)break;
		}
		if(cnt<4){
			for(map<int,int>::iterator it=mp.begin();it!=mp.end();it++){
				while((it->second)>=2){
					p[cnt++]=it->first;
					mp[it->first]-=2;
				}
				if(cnt>3)break;
			}
		}
		if(cnt<4){
			puts("NO");
		}
		else{
			puts("YES");
			if((p[2]-p[0])*(p[3]-p[1])>(p[2]-p[1])*(p[3]-p[0])){
				cout<<p[2]<<" "<<p[3]<<" "<<p[0]<<" "<<p[3]<<" "<<p[2]<<" "<<p[1]<<" "<<p[0]<<" "<<p[1]<<"\n";
			}
			else{
				cout<<p[2]<<" "<<p[3]<<" "<<p[1]<<" "<<p[3]<<" "<<p[2]<<" "<<p[0]<<" "<<p[1]<<" "<<p[0]<<"\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
1
8
1 1 1 1 2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
//MaximaApex
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string.h>
#include <numeric>
#include <map>
#include <set>

using namespace std;

#define ll long long
#define endl "\n" //comment out for interactive problems
const int MOD = 1e9+7;

ll roof(double x){return (ll)x == x ? x : (ll)x+1LL;}
ll gcd(ll a,ll b){return a==0 ? b : gcd(b%a,a);}
ll lcm(ll a,ll b){return a*(b/gcd(a,b));}

int main(){
    int tt;
    cin>>tt;
    while(tt--){
        int n;
        cin>>n;
        vector < int > arr(n);
        for(int i=0;i<n;i++)cin>>arr[i];

        map < int, int > mp;
        for(int i=0;i<n;i++){
            mp[arr[i]]++;
        }

        vector < pair < int, int > > brr;
        for(auto i : mp){
            if(i.second>=2)brr.push_back(make_pair(i.first,i.second));
        }

        if(brr.size()==1){
            if(brr[0].second>=8){
                cout<<"YES\n";
                for(int i=0;i<8;i++)cout<<brr[0].first<<" ";
                cout<<"\n";
            }
            else{
                cout<<"NO\n";
            }
            continue;
        }

        int x1,x2,y1,y2;
        if(brr[0].second>=4){
            x1 = brr[0].first;
            y1 = x1;
            brr[0].second-=4;
        }
        else{
            x1 = brr[0].first;
            y1 = brr[1].first;
            brr[0].second-=2;
            brr[1].second-=2;
        }
        if(brr.back().second>=4){
            y2 = brr.back().first;
            x2 = y1;
            brr.back().second = 0;
        }
        else if(brr.back().second>=2){
            y2 = brr.back().first;
            if(brr[brr.size()-2].second>=2){
                x2 = brr[brr.size()-2].first;
            }
            else{
                cout<<"NO\n";
                continue;
            }
        }
        else{
            cout<<"NO\n";
            continue;
        }

        cout<<"YES\n"<<x1<<" "<<y1<<" "<<x1<<" "<<y2<<" "<<x2<<" "<<y1<<" "<<x2<<" "<<y2<<"\n";

    }
}

**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
8
1 1 1 1 2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
t=int(input())
result=[]
for _ in range(t):
    n=int(input())
    nums=list(map(int,input().split()))
    updated_nums=[]
    for i in set(nums):
        if nums.count(i)>=2:
            updated_nums.append(i)
    updated_nums.sort()
    if len(updated_nums)<4:
        second_update=[]
        for i in set(updated_nums):
            if nums.count(i)>=4:
                second_update.append(i)
        if len(second_update)<2:
            third_update=[]
            for i in set(second_update):
                if nums.count(i)>=8:
                    third_update.append(i)
            if len(third_update)==0:
                if len(updated_nums)<2:
                    result.append('NO')
                else:
                    fourth_update=[]
                    for i in set(updated_nums):
                        if nums.count(i)>=6:
                            fourth_update.append(i)
                    if len(fourth_update)==0:
                        result.append('NO')
                    else:
                        #print(fourth_update)
                        for item in updated_nums:
                            if item != fourth_update[0]:
                                num=item
                                break
                        result.extend(['YES']+3*[fourth_update[0]]+[num])
                        result.extend(3*[fourth_update[0]]+[num])
            else:
                result.extend(['YES']+8*[third_update[0]])
        else:
            result.extend(['YES']+4*[second_update[0],second_update[1]])
    else:
        area1=(updated_nums[-1]-updated_nums[0])*(updated_nums[-2]-updated_nums[1])
        area2=(updated_nums[-1]-updated_nums[1])*(updated_nums[-2]-updated_nums[0])
        if area1>=area2:
            result.append('YES')
            result.extend([updated_nums[0],updated_nums[1],updated_nums[0],updated_nums[-2]])
            result.extend([updated_nums[-1],updated_nums[1],updated_nums[-1],updated_nums[-2]])
        else:
            result.append('YES')
            result.extend([updated_nums[0],updated_nums[-1],updated_nums[0],updated_nums[1]])
            result.extend([updated_nums[-2],updated_nums[-1],updated_nums[-2],updated_nums[1]])
for i in range(len(result)):
    if result[i] =='YES' or result[i]=='NO':
        print(result[i])
    else:
        if i+1<len(result) and (result[i+1]=='YES' or result[i+1]=='NO'):
            print(result[i])
        else:
            print(result[i],end=' ')

**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 focus on edge cases and scenarios that maximize the failure probability. Here are the steps to create such a test input:

1. **Edge Case with Few Unique Values**: The buggy code has multiple nested conditions to handle cases where there are few unique values. We can create a test case where there are exactly 8 elements, but only a few unique values to see how the code handles it.

2. **Edge Case with Repeated Values**: The code attempts to handle cases where certain values are repeated multiple times. We can create a test case where some values are repeated exactly 2, 4, 6, or 8 times to see how the code handles these conditions.

3. **Edge Case with Sorted Values**: The code sorts the values and then tries to form a rectangle. We can create a test case where the values are already sorted to see if the code correctly forms the rectangle.

4. **Edge Case with Negative Values**: The code should handle negative values correctly. We can create a test case with a mix of positive and negative values to see if the code handles them correctly.

Based on these considerations, here is a challenging test input:

```plaintext
3
8
1 1 1 1 1 1 1 1
8
1 2 3 4 5 6 7 8
8
-5 -5 0 0 5 5 10 10
```

### Explanation:
1. **Test Case 1**: All elements are the same (1). The code should print `YES` and a degenerate rectangle (area 0).
2. **Test Case 2**: All elements are distinct (1 to 8). The code should print `NO` because it's impossible to form a rectangle with these values.
3. **Test Case 3**: Mixed positive and negative values, with some values repeated. The code should print `YES` and the coordinates of a valid rectangle.

This test input should expose the bugs in the provided code by testing various edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
from collections import Counter
t = int(input())

for _ in range(t):
	n = int(input())
	a = list(map(int, input().split()))
	# print(a)
	c = Counter(a)
	v=[]
	for x in c:
		v.append((x,c[x]))
	# print(v)
	points = ["*"]*8
	e = -1
	for ind, i in enumerate(v):
		if points[0]=="*" and i[1]>=2:
			points[0]= str(i[0])
			points[2]=points[0]
		elif points[0]!="*" and i[1]>=2:
			points[1]=str(i[0])
			points[5]=points[1]
			e= ind
			break

	if e!=-1:
		# print("initial",e)
		# print(v)
		# print(e,v[:e:-1])
		for ind, i in enumerate(v[:e:-1]):
			if points[3]=="*" and i[1]>=2:
				points[3]=str(i[0])
				points[7]=points[3]
			elif points[3]!="*" and i[1]>=2:
				points[4]=str(i[0])
				points[6]=points[4]
				break
	# print(points)
	def isDone():
		for i in points:
			if i=="*":
				return False
		return True
	if(isDone()==False):
		vNew = sorted(v,key = lambda x:x[1],reverse=True)
		# print(vNew)
		if vNew[0][1]>=8:
			points= [vNew[0][0]]*8
		elif vNew[0][1]>=6 and vNew[1][1]>=2:
			points= [vNew[0][0]]*8
			points[3]=points[7]=vNew[1][0]
		elif vNew[0][1]>=4 and vNew[1][1]>=4:
			points=[]
			for i in range(4):
				points.append(vNew[0][0])
				points.append(vNew[1][0])

	if(isDone()==False):
		print("NO",end="")
	else:
		print("YES")
		for i in points:
			print(i,end=" ")
	print()

**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 create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that might cause the code to fail. Here are some considerations:

1. **Repeating Elements**: The code relies on finding multiple occurrences of the same element to form the coordinates. If there are not enough repeating elements, the code might fail to form a valid rectangle.

2. **Order of Elements**: The code uses the order of elements to form the coordinates, which might lead to incorrect results if the order is not considered correctly.

3. **Degenerate Rectangle**: The code should handle cases where a degenerate rectangle (area 0) is the best possible solution.

4. **Minimum and Maximum Values**: Using the minimum and maximum values in the list to form the coordinates might help in exposing bugs related to the selection of coordinates.

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

```plaintext
3
8
1 1 2 2 3 3 4 4
8
1 2 3 4 5 6 7 8
8
1 1 1 1 1 1 1 1
```

### Explanation:
1. **First Test Case**:
   - The list contains exactly 8 elements, each repeated twice.
   - This should form a valid rectangle with area 0, as all coordinates are the same.

2. **Second Test Case**:
   - The list contains 8 unique elements.
   - It is impossible to form a valid rectangle with these elements, so the output should be "NO".

3. **Third Test Case**:
   - The list contains 8 identical elements.
   - This should form a valid degenerate rectangle with area 0.

This test input should help in identifying issues in the buggy code, particularly in handling cases where there are not enough repeating elements or where all elements are the same.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
# لا تَجلِس فارِغًا [ حرك لسانك ]
# اسْتَغفِر ، سَبّح ، هَلّل ، كَبّر ، حَولق

import sys
from random import randint, shuffle, choice
from math import gcd , sqrt, isqrt, perm, comb, factorial, log2, ceil, floor
from collections import Counter, defaultdict, deque
from functools import lru_cache, reduce, cmp_to_key
from itertools import accumulate, combinations, permutations , product , repeat, takewhile , starmap , dropwhile , cycle
from heapq import nsmallest, nlargest, heappushpop, heapify, heappop, heappush , _heapify_max
from bisect import bisect_left, bisect_right , insort_left , insort_right
#from operator import sub , mul,pow  , truediv , floordiv
input = lambda: sys.stdin.buffer.readline().decode().rstrip()
OneByOne = lambda: sys.stdin.read(1)
def lcm(a , b) :
    return a * b // gcd(a ,b)
def W(i) :
    return (i , str(i))

def MakeFile(s : str) :
    import sys
    sys.stdin = open(f'{s}.in', 'r')
    sys.stdout = open(f'{s}.out', 'w')

I = lambda: input()
II = lambda: int(input())
MII = lambda: map(int, input().split())
LI = lambda: list(input().split())
LII = lambda: list(map(int, input().split()))
GMI = lambda: map(lambda x: int(x) - 1, input().split())
LGMI = lambda: list(map(lambda x: int(x) - 1, input().split()))
yes = lambda: print('YES')
no = lambda: print('NO')
DIR4 = ((-1, 0), (0, 1), (1, 0), (0, -1))
DIR8 = ((-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1))
MOD = 10 ** 9 + 7
inf = float('inf')

#for _ in range(II()) :

for _ in range(II()):
    n = II()
    l = LII()
    data = Counter(l)
    ref6 = []
    ref4 = []
    ref2 = []
    result = None

    for i, j in data.items():

        if j > 5:
            ref6.append(i); ref4.append(i); ref2.append(i)
        elif j > 3:
            ref4.append(i); ref2.append(i)
        elif j > 1:
            ref2.append(i)

        if j > 7 :
            result = [i] * 8

    area = 0
    ref6.sort() ; ref4.sort() ; ref2.sort()

    if ref6:
        for x in ref6:
            for j in ref2 :
                if x != j :
                    result = [x, x, x, j, x, x, x, j]

    if len(ref4) > 1:
        x = ref4[0] ; y = ref4[-1]
        if abs(x) * abs(y) > area :
            result = [x, x, x, y, y, x , y , y]
            area = abs(x) * abs(y)
    if len(ref2) > 3 :
            x1, y1 , y2 , x2 = ref2[0] , ref2[1], ref2[-1] , ref2[-2]
            if area < abs(y1 - y2) *abs(x1 - x2) :
                result = [x1 , y1 , x1, y2, x2, y1, x2, y2]

    if result:
        print("YES")
        print(*result)
    else:
        print("NO")

**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
8
1 1 1 1 1 1 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            HashMap<Integer,Integer> hm=new HashMap<>();
            int max,max2,min,min2;
            max=max2=Integer.MIN_VALUE;
            min=min2=Integer.MAX_VALUE;
            int n=sc.nextInt();
            for(int i=0;i<n;i++){
                int num=sc.nextInt();
                hm.put(num,hm.getOrDefault(num,0)+1);
                int f=hm.get(num);
                if(f%2==0){
                    if(min>num){
                        f-=2;
                        min2=min;
                        min=num;
                        hm.put(num,f);
                    }
                    else if(min2>num){
                        min2=num;
                        f-=2;
                        hm.put(num,f);
                    }
                }
                if(f>1 && f%2==0){
                    if(max<num){
                        max2=max;
                        max=num;
                        f-=2;
                        hm.put(num,f);
                    }
                    else if(max2<num){
                        max2=num;
                        f-=2;
                        hm.put(num,f);
                    }
                }
            }
            if(max2==Integer.MIN_VALUE || min2==Integer.MAX_VALUE)
                System.out.println("NO");
            else{
                System.out.println("YES");
                System.out.println(min+" "+min2+" "+min+" "+max+" "+max2+" "+min2+" "+max2+" "+max);
            }
        }

    }
}

**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 expose the bug in the provided code, we need to focus on the logic that handles the selection of the minimum and maximum values. The code seems to have issues with correctly identifying and updating the second smallest and second largest values, especially when there are multiple occurrences of the same value.

### Step-by-Step Analysis:
1. **Edge Case with Multiple Occurrences:**
   - The code should handle cases where there are multiple occurrences of the same value correctly. For example, if the input list contains multiple `0`s, the code should still be able to identify the second smallest and second largest values.

2. **Handling of `min2` and `max2`:**
   - The code updates `min2` and `max2` based on the frequency of the values. If the frequency is even, it tries to update `min2` and `max2` accordingly. However, this logic might fail if the values are not distinct.

3. **Degenerate Rectangle:**
   - The code should also handle cases where a degenerate rectangle (with area 0) is the best possible solution.

### Test Input:
Let's create a test input that includes multiple occurrences of the same value and ensures that the logic for `min2` and `max2` is tested:

```plaintext
3
12
0 0 0 0 0 0 0 0 0 0 0 0
12
0 0 0 0 1 1 1 1 2 2 2 2
12
-1 -1 -1 -1 1 1 1 1 2 2 2 2
```

### Explanation:
1. **Test Case 1:**
   - All values are `0`. This should result in a degenerate rectangle with all corners at `(0, 0)`.

2. **Test Case 2:**
   - Values are `0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2`. This should result in a rectangle with corners at `(0, 0)`, `(0, 2)`, `(2, 0)`, and `(2, 2)`.

3. **Test Case 3:**
   - Values are `-1, -1, -1, -1, 1, 1, 1, 1, 2, 2, 2, 2`. This should result in a rectangle with corners at `(-1, 1)`, `(-1, 2)`, `(2, 1)`, and `(2, 2)`.

This test input will help expose any issues with the logic for updating `min2` and `max2` and handling degenerate rectangles.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
import java.util.*;

public class contest2 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = Integer.parseInt(sc.nextLine());
        while(t>0){
            t--;
            int n = sc.nextInt();
            int[] arr = new int[n];
            for(int i=0; i<n; i++){
                arr[i] = sc.nextInt();
            }
            Arrays.sort(arr);
            Stack<Integer> st = new Stack<>();
            ArrayList<Integer> ar = new ArrayList<>();
            int count = 0;
            int temp = arr[0];
            for(int i=1; i<n; i++){
                if(arr[i]==temp) {
                    count+=2;
                }
                if(count==2) {
                    count=0;
                    st.push(arr[i]);
                    if(ar.size()==0 || ar.get(ar.size()-1)!=arr[i]){
                        ar.add(arr[i]);
                    }
                }
                temp=arr[i];
            }
            if(st.size()<4) System.out.println("NO");
            else{
                System.out.println("Yes");
                if(ar.size()>3){
                    int a1 = ar.get(0);
                    int a2 = ar.get(1);
                    int a3 = ar.get(ar.size()-1);
                    int a4 = ar.get(ar.size()-2);
                    if((a3-a1)*(a4-a2)>=(a4-a1)*(a3-a2)){
                        System.out.println(a1+" "+a2+" "+a1+" "+a4+" "+a3+" "+a2+" "+a3+" "+a4);
                    }
                    else System.out.println(a1+" "+a2+" "+a1+" "+a3+" "+a4+" "+a2+" "+a4+" "+a3);
                }
                else{
                    int a1 = st.pop();
                    int a2 = st.pop();
                    int a3 = st.pop();
                    int a4 = st.pop();
                    System.out.println(a1+" "+a2+" "+a1+" "+a4+" "+a3+" "+a2+" "+a3+" "+a4);
                }
            }
        }
    }
}

**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
8
1 1 1 1 1 1 1 1
8
0 0 0 0 1 1 1 1
8
-1 -1 2 2 3 3 4 4
```