**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Primary Task
time_limit_ms: 1000
memory_limit_mb: 256
Description: Dmitry wrote down $t$ integers on the board, and that is good. He is sure that he lost an important integer $n$ among them, and that is bad.

The integer $n$ had the form $\text{10^x}$ ($x \ge 2$), where the symbol '$\text{^}$' denotes exponentiation.. Something went wrong, and Dmitry missed the symbol '$\text{^}$' when writing the important integer. For example, instead of the integer $10^5$, he would have written $105$, and instead of $10^{19}$, he would have written $1019$.

Dmitry wants to understand which of the integers on the board could have been the important integer and which could not.

Input: The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of integers on the board.

The next $t$ lines each contain an integer $a$ ($1 \le a \le 10000$) — the next integer from the board.

Output: For each integer on the board, output "YES" if it could have been the important integer and "NO" otherwise.

You may output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
100
1010
101
105
2033
1019
1002
output:
NO
YES
NO
YES
NO
YES
NO

Note:

Buggy Code:
n=int(input())
for i in range(n):
      a=int(input())
      if a<100:
            print('NO')
      elif a>199 and a<999:
            print('NO')
      elif a>1999:
            print('NO')
      elif a<1000:
            if a%10==1 or a%10==0 or a>=200:
                  print('NO')
            else:
                  print('YES')
      elif a>1000:
            if a%100<10 or a>=2000:
                  print('NO')
            else:
                  print('YES')

**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 buggy code, we need to focus on edge cases and scenarios where the code might fail. The code has several conditions that might not handle all cases correctly. Let's break it down:

1. **Small Numbers (a < 100)**: The code correctly identifies that numbers less than 100 cannot be the important integer.
2. **Numbers between 200 and 999**: The code incorrectly identifies all numbers in this range as not being the important integer.
3. **Numbers between 1000 and 1999**: The code has a complex condition that might fail for certain numbers.
4. **Numbers greater than 1999**: The code incorrectly identifies all numbers greater than 1999 as not being the important integer.

To expose these issues, we can create a test input with numbers that fall into these critical ranges:

- Numbers less than 100
- Numbers between 200 and 999
- Numbers between 1000 and 1999
- Numbers greater than 1999

Here is a small-scale test input that covers these cases:

```plaintext
5
100
200
1010
1099
2000
```

This test input includes:
- 100: A number less than 100 (should be NO)
- 200: A number between 200 and 999 (should be NO, but the code might handle it incorrectly)
- 1010: A number between 1000 and 1999 (should be YES, but the code might handle it incorrectly)
- 1099: A number between 1000 and 1999 (should be NO, but the code might handle it incorrectly)
- 2000: A number greater than 1999 (should be NO, but the code might handle it incorrectly)

This should help expose the bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Primary Task
time_limit_ms: 1000
memory_limit_mb: 256
Description: Dmitry wrote down $t$ integers on the board, and that is good. He is sure that he lost an important integer $n$ among them, and that is bad.

The integer $n$ had the form $\text{10^x}$ ($x \ge 2$), where the symbol '$\text{^}$' denotes exponentiation.. Something went wrong, and Dmitry missed the symbol '$\text{^}$' when writing the important integer. For example, instead of the integer $10^5$, he would have written $105$, and instead of $10^{19}$, he would have written $1019$.

Dmitry wants to understand which of the integers on the board could have been the important integer and which could not.

Input: The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of integers on the board.

The next $t$ lines each contain an integer $a$ ($1 \le a \le 10000$) — the next integer from the board.

Output: For each integer on the board, output "YES" if it could have been the important integer and "NO" otherwise.

You may output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
100
1010
101
105
2033
1019
1002
output:
NO
YES
NO
YES
NO
YES
NO

Note:

Buggy Code:
for _ in range(int(input())):
    a=input()
    flag=False
    if a[0]=='1' and a[1]=='0' and (len(a)>2 and a[2]>='2' or len(a)>3 and a[2]=='1'):
        flag=True
    if flag:
        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
5
1000
1001
1010
1019
1020
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Primary Task
time_limit_ms: 1000
memory_limit_mb: 256
Description: Dmitry wrote down $t$ integers on the board, and that is good. He is sure that he lost an important integer $n$ among them, and that is bad.

The integer $n$ had the form $\text{10^x}$ ($x \ge 2$), where the symbol '$\text{^}$' denotes exponentiation.. Something went wrong, and Dmitry missed the symbol '$\text{^}$' when writing the important integer. For example, instead of the integer $10^5$, he would have written $105$, and instead of $10^{19}$, he would have written $1019$.

Dmitry wants to understand which of the integers on the board could have been the important integer and which could not.

Input: The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of integers on the board.

The next $t$ lines each contain an integer $a$ ($1 \le a \le 10000$) — the next integer from the board.

Output: For each integer on the board, output "YES" if it could have been the important integer and "NO" otherwise.

You may output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
100
1010
101
105
2033
1019
1002
output:
NO
YES
NO
YES
NO
YES
NO

Note:

Buggy Code:
import java.util.*;

public class deque {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
int t=sc.nextInt();
while(t-- >0){
   int a=sc.nextInt();
   Stack<Integer> st = new Stack<>();
   while(a>0){
       st.push(a%10);
       a=a/10;
   }
   if(st.pop()!=1){
       System.out.println("No");
       continue;
   }
    if(st.pop()!=0){
        System.out.println("No");
        continue;
    }
    if(st.size()==1 && st.peek()==1){
        System.out.println("No");
        continue;
    }
    if(st.pop()==0){
        System.out.println("No");
        continue;
    }

    System.out.println("Yes");
}

    }
}

**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
100
1010
101
1000
1001
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Primary Task
time_limit_ms: 1000
memory_limit_mb: 256
Description: Dmitry wrote down $t$ integers on the board, and that is good. He is sure that he lost an important integer $n$ among them, and that is bad.

The integer $n$ had the form $\text{10^x}$ ($x \ge 2$), where the symbol '$\text{^}$' denotes exponentiation.. Something went wrong, and Dmitry missed the symbol '$\text{^}$' when writing the important integer. For example, instead of the integer $10^5$, he would have written $105$, and instead of $10^{19}$, he would have written $1019$.

Dmitry wants to understand which of the integers on the board could have been the important integer and which could not.

Input: The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of integers on the board.

The next $t$ lines each contain an integer $a$ ($1 \le a \le 10000$) — the next integer from the board.

Output: For each integer on the board, output "YES" if it could have been the important integer and "NO" otherwise.

You may output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
100
1010
101
105
2033
1019
1002
output:
NO
YES
NO
YES
NO
YES
NO

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        int t = Integer.parseInt(br.readLine());
        for (int i = 0; i < t; i++) {
            String a= br.readLine();
            int secondone = Integer.parseInt(a.substring(2, a.length()));
            //System.out.println(secondone);
            if(a.charAt(0)=='1'&& a.charAt(1)=='0'){
                if(secondone>=2 && a.charAt(2)!='0')pw.println("YES");
                else pw.println("NO");
            }
            else pw.println("NO");
        }
        pw.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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The main issue in the buggy code is that it only checks if the first two characters are '10' and the third character is not '0', without properly validating the entire number.

Here are some edge cases to consider:
1. Numbers that start with '10' but are not in the form of '10^x'.
2. Numbers that start with '10' and are in the form of '10^x'.
3. Numbers that do not start with '10'.

Let's create a test input that includes these cases:

```plaintext
7
100
1010
101
105
1020
1019
1002
```

Explanation:
- `100` should be `NO` because it is not in the form `10^x` where `x >= 2`.
- `1010` should be `YES` because it could be `10^10`.
- `101` should be `NO` because it is not in the form `10^x` where `x >= 2`.
- `105` should be `YES` because it could be `10^5`.
- `1020` should be `NO` because it is not in the form `10^x` where `x >= 2`.
- `1019` should be `YES` because it could be `10^19`.
- `1002` should be `NO` because it is not in the form `10^x` where `x >= 2`.

This test input will help expose the bug in the provided code, which only checks the first few characters without validating the entire number.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Primary Task
time_limit_ms: 1000
memory_limit_mb: 256
Description: Dmitry wrote down $t$ integers on the board, and that is good. He is sure that he lost an important integer $n$ among them, and that is bad.

The integer $n$ had the form $\text{10^x}$ ($x \ge 2$), where the symbol '$\text{^}$' denotes exponentiation.. Something went wrong, and Dmitry missed the symbol '$\text{^}$' when writing the important integer. For example, instead of the integer $10^5$, he would have written $105$, and instead of $10^{19}$, he would have written $1019$.

Dmitry wants to understand which of the integers on the board could have been the important integer and which could not.

Input: The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of integers on the board.

The next $t$ lines each contain an integer $a$ ($1 \le a \le 10000$) — the next integer from the board.

Output: For each integer on the board, output "YES" if it could have been the important integer and "NO" otherwise.

You may output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
100
1010
101
105
2033
1019
1002
output:
NO
YES
NO
YES
NO
YES
NO

Note:

Buggy Code:
#include <bits/stdc++.h>
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")

using namespace std;

#define ll long long
#define ull unsigned long long

void answer(const bool& condition) {
    if (condition) {
        cout << "Yes\n";
    } else {
        cout << "No\n";
    }
}

inline void solve() {
    string a;
    cin >> a;
    if (a[0] == '1' && a[1] == '0') {
        if (a.size() == 3 && a[2] >= '2') {
            answer(true);
            return;
        } else if (a.size() == 3 && a[2] < '2') {
            answer(false);
            return;
        }
        if (a[2] == '0') {
            answer(false);
        } else {
            answer(true);
        }
        return;
    }
    answer(false);
}

int main() {
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios::sync_with_stdio(false);
    ull 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 given buggy code, we need to focus on edge cases that the code might handle incorrectly. The main issues in the code are:

1. **Incorrect handling of numbers with more than 3 digits**:
   - The code assumes that if the number starts with "10" and has a third digit that is not '0', it should be considered valid. However, it does not check if the remaining digits form a valid exponent.

2. **Incorrect handling of numbers with exactly 3 digits**:
   - The code checks if the third digit is greater than or equal to '2' but does not handle cases where the number is "101" or "100" correctly.

3. **Incorrect handling of numbers with leading zeros**:
   - The code does not handle leading zeros, which can cause issues if the input is not strictly formatted as expected.

Given these issues, let's create a small-scale test input that includes edge cases to expose these bugs:

```plaintext
7
100
101
102
1010
1011
1002
1099
```

Explanation:
- **100**: Should be "NO" because it cannot be written as 10^x.
- **101**: Should be "NO" because it cannot be written as 10^x.
- **102**: Should be "NO" because it cannot be written as 10^x.
- **1010**: Should be "YES" because it can be written as 10^10.
- **1011**: Should be "YES" because it can be written as 10^11.
- **1002**: Should be "NO" because it cannot be written as 10^x.
- **1099**: Should be "NO" because it cannot be written as 10^x.

This test input will help identify the issues in the buggy code, particularly in how it handles numbers with more than 3 digits and the specific edge cases for numbers starting with "10".


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Primary Task
time_limit_ms: 1000
memory_limit_mb: 256
Description: Dmitry wrote down $t$ integers on the board, and that is good. He is sure that he lost an important integer $n$ among them, and that is bad.

The integer $n$ had the form $\text{10^x}$ ($x \ge 2$), where the symbol '$\text{^}$' denotes exponentiation.. Something went wrong, and Dmitry missed the symbol '$\text{^}$' when writing the important integer. For example, instead of the integer $10^5$, he would have written $105$, and instead of $10^{19}$, he would have written $1019$.

Dmitry wants to understand which of the integers on the board could have been the important integer and which could not.

Input: The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of integers on the board.

The next $t$ lines each contain an integer $a$ ($1 \le a \le 10000$) — the next integer from the board.

Output: For each integer on the board, output "YES" if it could have been the important integer and "NO" otherwise.

You may output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
100
1010
101
105
2033
1019
1002
output:
NO
YES
NO
YES
NO
YES
NO

Note:

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

#define IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define rep(i,x,n) for(int i=x;i<n;++i)
#define rev(i,x,n) for(int i=n-1;i>=x;--i)
#define v(num) v.push_back(num)
#define ll long long
#define int(n) int n; cin>>n
#define Imax INT_MAX
#define Imin INT_MIN
#define all(x) x.begin(),x.end()

void solve(){

    string s;
    cin>>s;

    int n = s.size();

    bool f = 0;
    int x = 0;

    rep(i,0,n-2){
        if(s[i] == '1' && s[i+1] == '0' && (s[i+2] != '0' || (s[i+2] != '1' && i+3 == n))){
            f = 1;
            x = i+1;
            break;
        }
    }

    int y = 0;
    int a = 0, b = 1;
    if(f){
        for(int i=n-1;i>x;i--){
            a += ((s[i] - '0')*b);
            b *= 10;
        }
    }

    if(f && a > 1) cout<<"YES\n";
    else cout<<"NO\n";

}

int main(){
    IOS;

    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:
```plaintext
3
1000
1010
1001
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Primary Task
time_limit_ms: 1000
memory_limit_mb: 256
Description: Dmitry wrote down $t$ integers on the board, and that is good. He is sure that he lost an important integer $n$ among them, and that is bad.

The integer $n$ had the form $\text{10^x}$ ($x \ge 2$), where the symbol '$\text{^}$' denotes exponentiation.. Something went wrong, and Dmitry missed the symbol '$\text{^}$' when writing the important integer. For example, instead of the integer $10^5$, he would have written $105$, and instead of $10^{19}$, he would have written $1019$.

Dmitry wants to understand which of the integers on the board could have been the important integer and which could not.

Input: The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of integers on the board.

The next $t$ lines each contain an integer $a$ ($1 \le a \le 10000$) — the next integer from the board.

Output: For each integer on the board, output "YES" if it could have been the important integer and "NO" otherwise.

You may output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
100
1010
101
105
2033
1019
1002
output:
NO
YES
NO
YES
NO
YES
NO

Note:

Buggy Code:
import math

z = int(input())
for _ in range(z):

    n = int(input())

    s = [int(d) for d in str(n)]
    an = 0

    for i in range(0, len(s) - 1):
        if s[i] == 1 and s[i + 1] == 0:
            k = i
            an = 1
            break

    if an == 0:
        print("NO")
    else:
        if k + 2 >= len(s):
            print("NO")
        elif s[k + 2] == 0:
            print("NO")
        elif s[k + 2] == 1 and k + 2 == len(s) - 1:
            print("NO")
        else:
            print("YES")

**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 provided buggy code, we need to focus on edge cases and scenarios that the code might handle incorrectly. Here are the steps to think through the process:

1. **Identify the Bug**: The code checks for the presence of "10" in the number and then attempts to determine if the subsequent digits form a valid exponent. The main issues are:
   - It does not correctly handle cases where the number starts with "10" but is not followed by a valid exponent.
   - It does not handle cases where the number has multiple "10" sequences.

2. **Edge Cases to Consider**:
   - Numbers that start with "10" but are not followed by a valid exponent.
   - Numbers that contain "10" but have additional digits that should be invalid.
   - Numbers that are valid but might be incorrectly classified as invalid.

3. **Constructing the Test Input**:
   - Include numbers that start with "10" but are not valid (e.g., 1000, 10000).
   - Include numbers that contain "10" but have invalid subsequent digits (e.g., 10100, 10101).
   - Include valid numbers (e.g., 102, 1010).

Here is the test input that includes these edge cases:

```plaintext
5
1000
10100
10101
102
1010
```

Explanation:
- `1000`: Starts with "10" but is not a valid exponent (should be "NO").
- `10100`: Contains "10" but has invalid subsequent digits (should be "NO").
- `10101`: Contains "10" but has invalid subsequent digits (should be "NO").
- `102`: Valid (should be "YES").
- `1010`: Valid (should be "YES").

This test input should help expose the bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Primary Task
time_limit_ms: 1000
memory_limit_mb: 256
Description: Dmitry wrote down $t$ integers on the board, and that is good. He is sure that he lost an important integer $n$ among them, and that is bad.

The integer $n$ had the form $\text{10^x}$ ($x \ge 2$), where the symbol '$\text{^}$' denotes exponentiation.. Something went wrong, and Dmitry missed the symbol '$\text{^}$' when writing the important integer. For example, instead of the integer $10^5$, he would have written $105$, and instead of $10^{19}$, he would have written $1019$.

Dmitry wants to understand which of the integers on the board could have been the important integer and which could not.

Input: The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of integers on the board.

The next $t$ lines each contain an integer $a$ ($1 \le a \le 10000$) — the next integer from the board.

Output: For each integer on the board, output "YES" if it could have been the important integer and "NO" otherwise.

You may output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
100
1010
101
105
2033
1019
1002
output:
NO
YES
NO
YES
NO
YES
NO

Note:

Buggy Code:
t  = int(input())
for _ in range(t):
    n = input()
    r = "NO"
    for i in range(1,(len(n)-1)):
        if n[i] == "0" and n[i-1] == "1" and int(n[i+1:]) > 1 and n[i+1] != "0":
            r = "YES"
            break
    print(r)

**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
100
101
102
1010
1011
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Primary Task
time_limit_ms: 1000
memory_limit_mb: 256
Description: Dmitry wrote down $t$ integers on the board, and that is good. He is sure that he lost an important integer $n$ among them, and that is bad.

The integer $n$ had the form $\text{10^x}$ ($x \ge 2$), where the symbol '$\text{^}$' denotes exponentiation.. Something went wrong, and Dmitry missed the symbol '$\text{^}$' when writing the important integer. For example, instead of the integer $10^5$, he would have written $105$, and instead of $10^{19}$, he would have written $1019$.

Dmitry wants to understand which of the integers on the board could have been the important integer and which could not.

Input: The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of integers on the board.

The next $t$ lines each contain an integer $a$ ($1 \le a \le 10000$) — the next integer from the board.

Output: For each integer on the board, output "YES" if it could have been the important integer and "NO" otherwise.

You may output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
100
1010
101
105
2033
1019
1002
output:
NO
YES
NO
YES
NO
YES
NO

Note:

Buggy Code:
import java.util.*;
public class Main {
  public static void main(String[] args) {
    Scanner sc=new Scanner(System.in);
    int test=sc.nextInt();
    for(int i=0;i<=test-1;i++){
      int n=sc.nextInt();
      Stack<Integer> stack=new Stack<Integer>();
      int k=n;
      String p=" ";
      int count=0;
      while(n>10){
      k=n%10;
        stack.push(k);
        n=n/10;

      }
      if(n<10 && count==0){
        System.out.println("NO");
        count++;
      }
      int [] arr=new int[stack.size()];
     int s=0;
      while(!stack.isEmpty()){
        arr[s]=stack.peek();
        p=p+stack.pop();
        s++;

      }
      int l=0;
      l=Integer.parseInt(p.trim());
      if(arr[0]==0 && count==0){
        System.out.println("NO");
        count++;
      }
      if(l<2 && n==10 && arr[0]!=0 && count==0){
        System.out.println("NO");
        count++;
      }
      if(l>=2 && n==10 && arr[0]!=0 && count==0){
        System.out.println("YES");
      }

    }
    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:
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 are likely to cause the code to fail. Here are some key points to consider:

1. **Single-digit numbers**: The code should handle single-digit numbers correctly, but since the problem specifies that the important integer has the form \(10^x\) where \(x \ge 2\), single-digit numbers should not be considered.
2. **Numbers that do not match the pattern**: Numbers like 100, 101, 102, etc., should be correctly identified as not matching the pattern.
3. **Numbers that match the pattern**: Numbers like 1010, 1019, etc., should be correctly identified as matching the pattern.
4. **Edge cases**: Numbers that are close to the boundary conditions, such as 1000, 10000, and numbers just above or below these boundaries.

Given these considerations, let's create a small-scale test input that includes a mix of these cases:

```plaintext
5
100
1010
101
105
1000
```

This test input includes:
- 100: Should output "NO" (not a valid \(10^x\) where \(x \ge 2\)).
- 1010: Should output "YES" (valid \(10^10\)).
- 101: Should output "NO" (not a valid \(10^x\) where \(x \ge 2\)).
- 105: Should output "YES" (valid \(10^5\)).
- 1000: Should output "NO" (not a valid \(10^x\) where \(x \ge 2\)).

This input will help expose the bug in the buggy code by testing various cases that the code should handle correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Primary Task
time_limit_ms: 1000
memory_limit_mb: 256
Description: Dmitry wrote down $t$ integers on the board, and that is good. He is sure that he lost an important integer $n$ among them, and that is bad.

The integer $n$ had the form $\text{10^x}$ ($x \ge 2$), where the symbol '$\text{^}$' denotes exponentiation.. Something went wrong, and Dmitry missed the symbol '$\text{^}$' when writing the important integer. For example, instead of the integer $10^5$, he would have written $105$, and instead of $10^{19}$, he would have written $1019$.

Dmitry wants to understand which of the integers on the board could have been the important integer and which could not.

Input: The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of integers on the board.

The next $t$ lines each contain an integer $a$ ($1 \le a \le 10000$) — the next integer from the board.

Output: For each integer on the board, output "YES" if it could have been the important integer and "NO" otherwise.

You may output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
7
100
1010
101
105
2033
1019
1002
output:
NO
YES
NO
YES
NO
YES
NO

Note:

Buggy Code:
import java.util.*;

public class codeforce {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        for (int i = 0; i < n; i++) {
            int temp = sc.nextInt();
            String s = String.valueOf(temp);
            if (s.length() <=2) {
                System.out.println("NO");
            } else {
                int a = temp % (int)(Math.pow(10, s.length()-2));
                int b = temp / (int)(Math.pow(10, s.length()-2));
                if (a <=2 || b != 10) {
                    System.out.println("NO");
                } else {
                    System.out.println("YES");
                }
            }

        }

    }
}

**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
102
103
1010
1011
1019
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
/*In the name of Almighty Allah*/
#include<bits/stdc++.h>
using namespace std;

#define                 endl                        '\n'
#define                 speed                       ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define                 YES                         cout<<"YES"<<endl;
#define                 Yes                         cout<<"Yes"<<endl;
#define                 yes                         cout<<"yes"<<endl;
#define                 NO                          cout<<"NO"<<endl;
#define                 no                          cout<<"no"<<endl;
#define                 No                          cout<<"No"<<endl;
#define                 ll                          long long
#define                 frac(a)                     cout.unsetf(ios::floatfield); cout.precision(a); cout.setf(ios::fixed,ios::floatfield);//frac(4); cout<<a; 1.1234
#define                 f(i,pp,n)                   for(ll (i) = pp; i <n; i++)
#define                 rf(i,pp,n)                  for(ll (i) = pp; i >=n; i--)
#define                 fr(i,pp,n)                  for(ll (i) = pp; i <=n; i++)
#define                 ita                         vector<int>::iterator it; for(it=v.begin();it!=v.end();it++) { cout<<*it<<" "; } cout<<endl;
#define                 ita2(n)                     for(auto pri : (n)) { cout<<pri<<" " ;} cout<<endl;
#define                 ita2d(n)                    for( auto u : (n)){ for(auto x : u) { cout<<x<<" "; } cout<<endl;} // for( ll i = 0 ; i<vectorsize; i++) { for( ll j = 0 ; j< v[i].size() ; j++) { cout << v[i][j]<<" "; }  cout<<endl; }
#define                 ita3(n)                     for(auto pri : (n)) { cout<<pri.first<<" "<<pri.second<<endl; }
#define                 vsort(n)                    sort(n.begin(),n.end());
#define                 vrsort(n)                   sort(n.begin(),n.end(),greater <> ());
#define                 MOD                         1000000007
#define                 bug                         cout<<"ekane asche"<<endl;
#define                 testcases                   ll t; cin >> t; while (t--)
#define                 unique(n)                   (n).erase(unique((n).begin(),(n).end()),(n).end());
#define                 mem(a,b)                    memset(a, b, sizeof(a) )
const long long inf = 90000000000000000;
ll test ,cnt = 0 ;
// // modulo // PBDS // graph_index // rotator // issort? // sum // count

void solve()
{
    ll n , m ,k;
    string s ; cin>>s;
    string st ; cin>>st;
    vector<ll>ps;
    f(i,0,st.size())
    {
        int f = 0  ;
        // cout<<st[i]<<endl;
        f(j,0,s.size())
        {
            if(s[j]==st[i] && (ps.size()==0 || ps.back()<j))
            {
                ps.push_back(j);
                f=1;
                break;
            }
        }
        if(!f && ps.size())
        {
            cout<<(st.size()-ps.size())+s.size()<<endl;
            return;
        }
    }
    cout<<(st.size()-ps.size())+s.size()<<endl;

}

int main()
{
    speed

    ll  c = 1;
    cin >> test;
    cnt = 0 ;
    for(ll tc = 1 ; tc<= test ; tc++ )
    {
        cnt++ ;
        // cin.ignore();
        // cout << "Case " << c++ << ": ";
        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:
```plaintext
3
abc
def
a
a
abc
cba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define vll vector<long long int>
#define fr(i,n) for(int i=0;i<n;i++)
int main(){
    ll t;
    cin>>t;
    while(t--){
        string a,b;
        cin>>a;
        cin>>b;
        vll v;
        map<int,bool>m1;
        fr(i,a.size()){
            m1[i]=false;
        }
        fr(i,b.size()){
            fr(j,a.size()){
                if(b[i]==a[j]){
                    if(!m1[j]){
                    v.push_back(j);
                    m1[j]=true;
                    break;
                    }else{
                        continue;
                    }

                }
            }
        }
        vll l(a.size(),0);
        ll c=0;
        if(v.size()==0){
            cout<<a.size()+b.size()<<endl;
        }
        else{
        fr(i,v.size()-1){
            if(v[i+1]>v[i]){
                l[c]++;
            }
            else{
                c++;
            }
        }
        cout<<a.size()+b.size()-*(max_element(l.begin(),l.end()))-1<<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
3
abc
cba
abac
bac
abcde
edcba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

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

using ll = long long;
using ull = unsigned long long;

using namespace std;

void solve() {
    string a, b;
    cin >> a >> b;

    int n_a = a.size();
    int n_b = b.size();
    int mx = 0;

    vector<vector<int>> dp(n_a, vector<int>(n_b));
    for (int i = 0; i < n_a; i++) {
        for (int j = 0; j < n_b; j++) {
            if (j > 0) {
                dp[i][j] = max(dp[i][j], dp[i][j-1]);
            }
            if (i > 0) {
                dp[i][j] = max(dp[i][j], dp[i-1][j]);
            }
            if (i > 0 && j > 0) {
                dp[i][j] = max(dp[i][j], dp[i-1][j-1]);
            }
            if (a[i] == b[j]) {
                dp[i][j] = max(dp[i][j], 1 + ((i > 0 && j > 0) ? dp[i-1][j-1] : 0));
            }
            mx = max(mx, dp[i][j]);
        }
    }

    cout << n_a + n_b - mx << "\n";
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	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:
```plaintext
1
abac
cba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
t = int(input())

for _ in range(t):
    a = input()
    b = input()

    maxSubsequence = 0
    fin_maxSubsequence = 0

    for i in range(len(b)):
        for j in range(len(a)):
            if b[i] == a[j]:
                maxSubsequence+=1
                if (j+1<=len(a) and i+1<=len(b)):
                    for x in range(i+1, len(b)):
                        for k in range(j+1, len(a)):
                            if a[k] == b[x]:
                                maxSubsequence+=1
                                x+=1

                                break
            fin_maxSubsequence = max(maxSubsequence, fin_maxSubsequence)
            maxSubsequence = 0

    print(len(a)+len(b)-fin_maxSubsequence)

**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
abc
cba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
for _ in range(int(input())):
    a=input()
    b=input()
    def longestCommonSubsequence( a, b):
        m = len(a)
        n = len(b)
        dp = [[""] * (n + 1) for _ in range(m + 1)]

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if a[i - 1] == b[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + a[i - 1]
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1], key=len)
        return len(dp[-1][-1])

    print(len(a)+len(b)-longestCommonSubsequence(a,b))

**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
abc
cba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
import sys
import math
import random
from heapq import *
from collections import Counter, defaultdict, deque
from functools import lru_cache
from itertools import permutations, combinations
from decimal import Decimal
from bisect import bisect_right, bisect_left, bisect
import re

input = lambda: sys.stdin.readline().strip()
minput = lambda: map(int, input().split())
linput = lambda: list(minput())
MOD = 1000000007
MOD2 = 998244353
INF = 10 ** 50
primes_N = 0
primes = [1]*(primes_N+1)
primes100 = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97)
primes1000 = (
    2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109,
    113,
    127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241,
    251,
    257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389,
    397,
    401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547,
    557,
    563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691,
    701,
    709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859,
    863,
    877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997)

def factorization_list(n):
    # Разложение числа на простые множители в виде списка чисел
    res = []
    while n % 2 == 0:
        n //= 2
        res.append(2)

    i = 3
    while i**2 <= n:
        while n % i == 0:
            n //= i
            res.append(i)
        i += 2
    if n > 1:
        res.append(n)
    return sorted(res)

def factorization_dict(n):
    # Разложение числа на простые множители в словарь
    res = defaultdict(int)
    while n % 2 == 0:
        res[2] += 1
        n //= 2
    i = 3
    while i**2 <= n:
        while n % i == 0:
            res[i] += 1
            n //= i
        i += 2
    if n > 1:
        res[n] += 1
    return res

def sieve():
    # Решето Эратосфена
    primes[0] = primes[1] = 0
    for j in range(4, primes_N+1, 2):
        primes[j] = 0

    i = 3
    while i**2 <= primes_N:
        if primes[i]:
            for j in range(i**2, primes_N+1, i):
                primes[j] = 0
        i += 2

def solve():
    a = input()
    b = input()
    n = len(a)
    m = len(b)
    k = 0
    left = 0
    for i in range(n):
        for j in range(left, m):
            if a[i] == b[j]:
                k += 1
                left = j
                break

    print((n-k)+(m-k)+k)

if __name__ == '__main__':
    tc = 1
    tc = int(input())
    for _ in range(tc):
        solve()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code has a logic that tries to find the minimum possible length of a string that contains `a` as a substring and `b` as a subsequence. However, it seems to have issues with correctly handling overlaps and the order of characters.

### Step-by-Step Analysis

1. **Identify the Bug**:
   - The code iterates through each character of `a` and tries to find the same character in `b` starting from the last found position.
   - It calculates the number of common characters (`k`) and then computes the result as `(n - k) + (m - k) + k`.

2. **Edge Cases to Consider**:
   - **Case 1**: `a` and `b` are identical.
   - **Case 2**: `a` is a substring of `b`.
   - **Case 3**: `b` is a subsequence of `a`.
   - **Case 4**: `a` and `b` have no common characters.
   - **Case 5**: `a` and `b` have some common characters, but the order is important.

3. **Generate Test Input**:
   - We need a test case that will show the bug clearly. A good candidate is a case where `a` and `b` have some common characters but the order is such that the code's logic fails to correctly compute the minimum length.

### Chosen Test Input
Let's consider the following test input:

- `a = "abc"`
- `b = "cba"`

### Explanation
- **a** is "abc".
- **b** is "cba".
- The minimum possible length of a string that contains "abc" as a substring and "cba" as a subsequence is 5. One such string is "abcba".

However, the buggy code will likely fail to correctly handle the order of characters and might compute an incorrect length.

### Output
```plaintext
1
abc
cba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
import java.util.*;

public class A{
    public static void main(String[] args) {

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

        while(t>0){
          String a=sc.next();
          String b=sc.next();
          String c="";
          StringBuilder str=new StringBuilder(b);
          for(int i=0;i<b.length();i++){
            String z=Character.toString(b.charAt(i));
            if(!a.contains(z)){
            c=c+b.charAt(i);
            }
            else{
                continue;
            }
          }
          System.out.println(a.length()+c.length());

            t--;
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
abc
abc
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
import java.util.*;

public class qu2 {
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        sc.nextLine();
        while(t-- > 0){
            String a = sc.nextLine();
            String b = sc.nextLine();

            int arr[] = new int[a.length()];
            Arrays.fill(arr, -1);
            for(int i = 0; i < b.length(); i++){
                char c = b.charAt(i);
                for(int j = 0; j < a.length(); j++){
                    if(a.charAt(j) == c && arr[j] == -1){
                        arr[j] = i;
                        break;
                    }
                }
            }

            List<Integer> list = new ArrayList<>();
            for(int i = 0; i < a.length(); i++){
                if(arr[i] != -1){
                    list.add(arr[i]);
                }
            }

            int lisLength = findLISLength(list);

            System.out.println(a.length() + b.length() - lisLength);
        }
        sc.close();
    }

    private static int findLISLength(List<Integer> list) {
        if(list.isEmpty()) return 0;

        int[] lis = new int[list.size()];
        int len = 1;
        lis[0] = list.get(0);

        for(int i = 1; i < list.size(); i++){
            if(list.get(i) > lis[len - 1]){
                lis[len++] = list.get(i);
            } else {
                int pos = Arrays.binarySearch(lis, 0, len, list.get(i));
                if (pos < 0) pos = -(pos + 1);
                lis[pos] = list.get(i);
            }
        }

        return len;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider scenarios where the code might fail. The code is supposed to find the minimum possible length of a string that contains `a` as a substring and `b` as a subsequence. The bug in the code is in how it handles the mapping of characters from `b` to `a` and the calculation of the longest increasing subsequence (LIS).

Here are the steps to think through the problem:

1. **Edge Cases:**
   - Both `a` and `b` are the same.
   - `a` is a substring of `b`.
   - `b` is a subsequence of `a`.
   - `a` and `b` have no common characters.
   - `a` and `b` have some common characters but in a way that the code might not correctly map them.

2. **Scenarios to Test:**
   - When `a` and `b` have multiple common characters in different positions.
   - When `b` has characters that are not in `a`.
   - When `a` and `b` have overlapping characters but in a way that the LIS calculation might fail.

3. **Challenging Input:**
   - Let's consider a case where `a` and `b` have some common characters, but the order in `b` is such that the code might incorrectly map them.

Given these considerations, a challenging test input could be:

```plaintext
1
abac
cba
```

### Explanation:
- **String `a`:** `abac`
- **String `b`:** `cba`

In this case:
- The string `a` is `abac`.
- The string `b` is `cba`.

The correct answer should be `4` because the string `cbaa` contains `a` as a substring and `b` as a subsequence.

However, the buggy code might fail to correctly map the characters from `b` to `a` and might not correctly calculate the LIS, leading to an incorrect result.

This input is small-scale and complete, and it should expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
/* "EXPLORE THE WORLD BIT BY BIT" */
                                              /* -TOWER_OF_HAMIM- */
                                           //IMAGINATION_IS_EVERYTHING//
#include <bits/stdc++.h>
using namespace std;

#define yes cout << "YES" << endl;
#define no cout << "NO" << endl;
#define End cout << '\n';
#define sp " "
#define ok continue;
#define mod 1000000007
#define md 1000000009
#define long long long int

long hamim( string &a,  string &b, long m, long n) {
    if (m < 0 or n < 0) {
        return 0;
    }
    if (a[m] == b[n]) {
        return 1 + hamim(a, b, m - 1, n - 1);
    } else {
        return max(hamim(a, b, m, n - 1), hamim(a, b, m - 1, n));
    }
}

void TOWER_OF_HAMIM(long tt) {
    while (tt--) {
        string a, b;
        cin >> a >> b;
        long m = a.size();
        long n = b.size();
        long x = hamim(a, b, m - 1, n - 1);
        long ans = m + n - x ;
        cout << ans << endl;
    }
}

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

    long t;
    cin >> t;
    TOWER_OF_HAMIM(t);
    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
abc
cba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
#include <bits/stdc++.h>
#include <iostream>
#include <string>
#define IAMFAST ios_base::sync_with_stdio(false);cin.tie(nullptr);
#define ll long long
#define el std::cout<<endl
#define fpi(s,e) for(int i=s;i<e;i++)
#define fip(s,e) for(int i=s;i>=e;i--)
#define fpj(s,e) for(int j=s;j<e;j++)
#define all(v) (v).begin(), (v).end()
#define MOD 1e9 + 7
using namespace std;
const int N = 1e6 + 5;

void solve(){
    string a, b;
    cin >> a >> b;
    int i = 0, ans = 0;
    int lastFound = 0;
    for(int j = 0; j < b.size(); j++){
        bool found = false;
        for(; i < a.size(); i++){
            if(a[i] == b[j]){
                found = true;
                break;
            }
        }
        if(i > a.size())
            ans++;
        else if(!found){
            i = lastFound;
            ans++;
        }else{
            lastFound = ++i;
        }
    }
    cout << ans + a.size();
}

int main(){
    IAMFAST
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; i++) {
        //cout<<"Case #"<<i<<": ";
        solve();
        if(i!=t){
            el;
        }
    }
    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 code, let's analyze the logic and potential edge cases:

1. **Bug Analysis:**
   - The code uses a nested loop to find the characters of string `b` in string `a`.
   - If a character from `b` is not found in `a`, the code resets the search position to `lastFound` and increments the answer.
   - The bug likely occurs when the search for a character in `b` fails, and the reset position is not correctly handled.

2. **Edge Cases to Consider:**
   - **String `a` contains characters from `b` but in a different order:** This can cause the search to reset incorrectly.
   - **String `a` is a prefix of `b` or vice versa:** This can cause the search to fail if the reset position is not handled correctly.
   - **String `a` and `b` have no common characters:** This should result in the length of `a` plus the length of `b`.

3. **Challenging Test Input:**
   - Let's create a test input where `a` and `b` have some overlapping characters, but the order is such that the search will fail and reset incorrectly.

Here is a small-scale test input that should expose the bug:

```plaintext
1
abc
cba
```

**Explanation:**
- For the input `a = "abc"` and `b = "cba"`, the correct answer should be 5. The string "abcba" contains "abc" as a substring and "cba" as a subsequence.
- The provided code will likely fail because it will reset the search position incorrectly when it fails to find the next character in `b` in `a`.

This test input should help identify the issue in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
def lcs(X, Y):
    # 두 문자열의 길이
    m = len(X)
    n = len(Y)

    # DP 테이블 초기화
    L = [[None]*(n+1) for i in range(m+1)]

    # DP 테이블을 채우기
    for i in range(m+1):
        for j in range(n+1):
            if i == 0 or j == 0:
                L[i][j] = 0
            elif X[i-1] == Y[j-1]:
                L[i][j] = L[i-1][j-1] + 1
            else:
                L[i][j] = max(L[i-1][j], L[i][j-1])

    # LCS의 길이
    length_of_lcs = L[m][n]
    return length_of_lcs

def solve():
    a = input()
    b = input()
    ans = len(a)
    ans += len(b) - lcs(a, b)
    print(ans)

T = int(input())
while T > 0:
    solve()
    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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases that can cause the algorithm to fail. The main issue in the buggy code is that it does not handle cases where the string `a` and string `b` have overlapping characters correctly. Specifically, the code might overestimate the length of the resulting string by not accounting for the characters that can be shared between `a` and `b`.

Here are some edge cases to consider:
1. **Complete Overlap**: `a` and `b` are the same string.
2. **Partial Overlap**: `a` and `b` share some characters but are not identical.
3. **No Overlap**: `a` and `b` have no common characters.
4. **Complex Overlap**: `a` and `b` have a complex pattern of shared characters that can be tricky to handle.

Given these considerations, let's create a test case that exposes the bug:

```plaintext
3
abc
abc
abc
def
abc
xyz
```

### Explanation:
1. **Complete Overlap**:
   - `a = "abc"`, `b = "abc"`
   - The correct output should be 3, as the string "abc" itself contains both `a` as a substring and `b` as a subsequence.

2. **Partial Overlap**:
   - `a = "abc"`, `b = "def"`
   - The correct output should be 6, as the string "abcdef" contains `a` as a substring and `b` as a subsequence.

3. **No Overlap**:
   - `a = "abc"`, `b = "xyz"`
   - The correct output should be 6, as the string "abcxyz" contains `a` as a substring and `b` as a subsequence.

This test case will help expose any issues in the buggy code, particularly in how it handles overlapping characters and the calculation of the minimum length of the resulting string.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
t = int(input())

for _ in range(t):
    a = input()
    b = input()

    maxSubsequence = 0
    fin_maxSubsequence = 0

    for i in range(len(b)):
        for j in range(len(a)):
            if b[i] == a[j]:
                maxSubsequence+=1
                if (i+1<len(b) and j+1<len(a)):
                    i+=1
        fin_maxSubsequence = max(maxSubsequence, fin_maxSubsequence)
        maxSubsequence = 0
    print(len(a)+len(b)-fin_maxSubsequence)

**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
abc
abcabc
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

Buggy Code:
import java.util.Scanner;

public class MinimumStringContainingSubstringAndSubsequence {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        scanner.nextLine();  // Consume the newline character

        for (int i = 0; i < t; i++) {
            String a = scanner.nextLine();
            String b = scanner.nextLine();
            System.out.println(findMinLength(a, b));
        }

        scanner.close();
    }

    private static int findMinLength(String a, String b) {
        StringBuilder result = new StringBuilder(a);
        int n = a.length();

        for (int i = 0; i < b.length(); i++) {
            char bChar = b.charAt(i);
            if (!containsChar(a, bChar)) {
                result.append(bChar);
            }
        }

        return result.length();
    }

    private static boolean containsChar(String s, char c) {
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == c) {
                return true;
            }
        }
        return false;
    }
}

**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
abc
def
abc
cba
abc
abc
abc
abcd
abc
abcde
```

5
abc
def
abc
cba
abc
abc
abc
abcd
abc
abcde


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

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

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            if (System.getProperty("ONLINE_JUDGE") == null) {
                try {
                    this.br = new BufferedReader(new FileReader("input.txt"));
                } catch (FileNotFoundException e) {
                    e.printStackTrace();
                }
            } else {
                this.br = new BufferedReader(new InputStreamReader(System.in));
            }

        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    String str = br.readLine();
                    if (str == null) {
                        return null; // End of input
                    }
                    st = new StringTokenizer(str);
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() { return Integer.parseInt(next()); }
        long nextLong() { return Long.parseLong(next()); }
        double nextDouble() { return Double.parseDouble(next()); }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter() {
            BufferedWriter writer;
            if (System.getProperty("ONLINE_JUDGE") == null) {
                try {
                    writer = new BufferedWriter(new FileWriter("output.txt"));
                } catch (IOException e) {
                    e.printStackTrace();
                    writer = new BufferedWriter(new OutputStreamWriter(System.out));
                }
            } else {
                writer = new BufferedWriter(new OutputStreamWriter(System.out));
            }
            this.bw = writer;
        }

        public void print(Object object) throws IOException {
            bw.append(String.valueOf(object));
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

    static long gcd(long a, long b) {
        while (b > 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    static long lcm(long a, long b) {
        return (a * b) / gcd(a, b);
    }

    static int[] prefix_sum(int[] arr) {
        int n = arr.length;
        int[] pref = new int[n];
        pref[0] = arr[0];
        for(int i = 1; i < n; i++)
            pref[i] = pref[i-1] + arr[i];
        return pref;
    }

    static long power(long x, long y, long p) {
        long res = 1;
        x = x % p;
        while (y > 0) {
            if ((y & 1) == 1)
                res = (res * x) % p;
            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }
    static class Sieve {
        boolean[] isPrime;
        int[] primes;
        int primeCount;
        int maxN;

        public Sieve(int n) {
            maxN = n;
            isPrime = new boolean[n + 1];
            Arrays.fill(isPrime, true);

            isPrime[0] = isPrime[1] = false;

            for (int i = 2; i * i <= n; i++) {
                if (isPrime[i]) {
                    for (int j = i * i; j <= n; j += i) {
                        isPrime[j] = false;
                    }
                }
            }
            countAndStorePrimes();
        }

        private void countAndStorePrimes() {
            primeCount = 0;
            for (int i = 2; i <= maxN; i++) {
                if (isPrime[i]) {
                    primeCount++;
                }
            }

            primes = new int[primeCount];
            int index = 0;
            for (int i = 2; i <= maxN; i++) {
                if (isPrime[i]) {
                    primes[index++] = i;
                }
            }
        }

        public boolean isPrime(int num) {
            if (num <= maxN) return isPrime[num];
            if (num <= 1) return false;
            for (int i = 2; i * i <= num; i++) {
                if (num % i == 0) return false;
            }
            return true;
        }

        public int[] getPrimes() {
            return primes;
        }

        public int getPrimeCount() {
            return primeCount;
        }

        public Map<Integer, Integer> getPrimeFactorization(int n) {
            Map<Integer, Integer> factors = new HashMap<>();

            for (int prime : primes) {
                if (prime * prime > n) break;

                while (n % prime == 0) {
                    factors.put(prime, factors.getOrDefault(prime, 0) + 1);
                    n /= prime;
                }
            }
            if (n > 1) {
                factors.put(n, 1);
            }

            return factors;
        }
    }

    static final int MOD = 1_000_000_007;
    static final int INF = Integer.MAX_VALUE;
    static final int MINI = Integer.MIN_VALUE;
    static final long LINF = Long.MAX_VALUE;
    static FastReader in;
    static FastWriter out;

    public static void solve() throws IOException {
        String a = in.nextLine();
        String b = in.nextLine();

        int ans = Math.max(a.length(),b.length());
        if(ans == a.length()){
            for(int i = 0 ; i < b.length();++i){
                char ch = b.charAt(i);
                if(a.indexOf(ch)==-1){
                    ans++;
                }
            }
        }else{
            for(int i = 0 ; i < a.length();++i){
                char ch = a.charAt(i);
                if(b.indexOf(ch)==-1){
                    ans++;
                }
            }
        }

        out.println(ans);
    }

    public static void main(String[] args) {
        try {
            in = new FastReader();
            out = new FastWriter();

            int t = 1;
            String input = in.next();
            if (input != null) {
                t = Integer.parseInt(input);
            }

            while (t-- > 0) {
                solve();
            }

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    static void debug(Object... args) {
        System.err.println(Arrays.deepToString(args));
    }

    static <T> List<T> asList(T[] arr) {
        return Arrays.asList(arr);
    }

    static int[] sort(int[] arr) {
        Arrays.sort(arr);
        return arr;
    }

    static int upperBound(int[] arr, int x) {
        int l = 0, r = arr.length;
        while (l < r) {
            int m = l + (r - l) / 2;
            if (arr[m] <= x) l = m + 1;
            else r = m;
        }
        return l;
    }

    static int lowerBound(int[] arr, int x) {
        int l = 0, r = arr.length;
        while (l < r) {
            int m = l + (r - l) / 2;
            if (arr[m] >= x) r = m;
            else l = m + 1;
        }
        return l;
    }
}

**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 where the code might fail. The buggy code logic is flawed in how it calculates the minimum possible length of the string that contains \(a\) as a substring and \(b\) as a subsequence. Specifically, the code only checks for the presence of characters from one string in the other, which can lead to incorrect results in certain scenarios.

### Step-by-Step Analysis:

1. **Identify the Bug:**
   - The code assumes that if the length of \(a\) is greater than or equal to the length of \(b\), it only checks if characters of \(b\) are in \(a\).
   - If the length of \(b\) is greater, it checks if characters of \(a\) are in \(b\).
   - This approach fails when the characters of \(b\) are not in \(a\) but can still form a subsequence in a combined string.

2. **Construct a Test Case:**
   - We need a test case where \(a\) and \(b\) have different lengths and the characters of \(b\) are not all present in \(a\), but the combined string can still be shorter than the sum of the lengths of \(a\) and \(b\).

### Example Test Case:
- Let \(a = "abc"\) and \(b = "def"\).
- The combined string "abcdef" contains \(a\) as a substring and \(b\) as a subsequence.
- The expected output should be 6, but the buggy code will incorrectly output 9 because it will add the length of \(b\) to the length of \(a\) since none of the characters of \(b\) are in \(a\).

### Challenging Test Input:
```plaintext
1
abc
def
```

### Explanation:
- The input consists of one test case.
- \(a = "abc"\) and \(b = "def"\).
- The minimum possible length of the string that contains \(a\) as a substring and \(b\) as a subsequence is 6.
- The buggy code will incorrectly output 9, exposing the bug in the logic.

This test input should effectively expose the flaw in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring and Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two strings $a$ and $b$, both consisting of lowercase Latin letters.

A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.

For example, consider the string abac:

  * a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences;    * a, b, c, ab, ba, ac, aba, bac and abac are its substrings.

Your task is to calculate the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^3$) — the number of test cases.

The first line of each test case contains a string $a$ ($1 \le |a| \le 100$), consisting of lowercase Latin letters.

The second line of each test case contains a string $b$ ($1 \le |b| \le 100$), consisting of lowercase Latin letters.

Output: For each test case, print a single integer — the minimum possible length of the string that contains $a$ as a substring and $b$ as a subsequence.

Examples:
input:
5
aba
cb
er
cf
mmm
mmm
contest
test
cde
abcefg
output:
4
4
3
7
7

Note: In the examples below, the characters that correspond to the subsequence equal to $b$ are bolded.

In the first example, one of the possible answers is caba.

In the second example, one of the possible answers is ercf.

In the third example, one of the possible answers is mmm.

In the fourth example, one of the possible answers is contest.

In the fifth example, one of the possible answers is abcdefg.

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

int lcs(string text1, string text2) {
        int len1 = text1.length(), len2 = text2.length();
        int cur_val, pre_val;
        vector<int> dp(len2+1, 0);
        for(int i=1;i<=len1;++i) {
            pre_val = 0;
            for(int j=1;j<=len2;++j) {
                cur_val = dp[j];
                dp[j] = max({dp[j], dp[j-1], pre_val + (text1[i-1]==text2[j-1])});
                pre_val = cur_val;
            }
        }
        return dp[len2];
    }

int main() {
    int t; cin>>t;
    while(t--){
        string a,b; cin>>a>>b;
        int ans = lcs(a,b),n=a.size(),m=b.size();
        cout<<n+m-ans<<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:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that might cause the code to fail. The code uses the Longest Common Subsequence (LCS) to calculate the minimum possible length of a string that contains both `a` as a substring and `b` as a subsequence. The main issue to consider is how the LCS is calculated and how it affects the final result.

### Key Points to Consider:
1. **Empty or Very Short Strings**: These can sometimes cause edge cases that are not handled well.
2. ** Strings with Repeated Characters**: These can affect the LCS calculation in non-intuitive ways.
3. ** Strings with No Common Subsequence**: This can cause the LCS to be zero, which might not be handled correctly.
4. ** Strings with a Lot of Common Characters**: This can cause the LCS to be large, which might not be handled correctly.

### Challenging Test Input:
Let's create a test case where `a` and `b` have a lot of common characters, but the LCS is not straightforward to compute. This can help expose any issues in the LCS calculation.

```plaintext
3
abac
baca
abc
cba
aaaa
aaaa
```

### Explanation:
1. **Test Case 1: `abac` and `baca`**
   - The LCS of `abac` and `baca` is `aba` (length 3).
   - The minimum length of the string that contains `abac` as a substring and `baca` as a subsequence is `5` (e.g., `abacac`).

2. **Test Case 2: `abc` and `cba`**
   - The LCS of `abc` and `cba` is `abc` (length 3).
   - The minimum length of the string that contains `abc` as a substring and `cba` as a subsequence is `6` (e.g., `abcba`).

3. **Test Case 3: `aaaa` and `aaaa`**
   - The LCS of `aaaa` and `aaaa` is `aaaa` (length 4).
   - The minimum length of the string that contains `aaaa` as a substring and `aaaa` as a subsequence is `4` (e.g., `aaaa`).

This test input should help expose any issues in the LCS calculation and the final result computation.