**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

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

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
import sys
sys.setrecursionlimit(2*(10**3))
for _ in range(int(input())):
    a = input()
    b = input()
    c = input()
    dic = {}
    dic = {}
    key = lambda i, j, k: i * len(b) * len(c) + j * len(c) + k
    def calculate(i,j,k,dic):
        key_ = key(i,j,k)
        if (i,j,k) in dic:
            return dic[key_]
        if i == len(a):
            count = 0
            while k <len(c):
                if c[k] != b[j]:
                    count+=1
                k+=1
                j+=1
            dic[key_] = count
            return count
        if j == len(b):
            count = 0
            while k < len(c):
                if c[k] != a[i]:
                    count+=1
                i+=1
                k+=1
            dic[key_] = count
            return count

        if c[k]!=a[i] and c[k]!=b[j]:
            count = 1 + min(calculate(i+1,j,k+1,dic),calculate(i,j+1,k+1,dic))
        elif c[k]!=a[i] and c[k] == b[j]:
            count = min(calculate(i,j+1,k+1,dic),calculate(i+1,j,k+1,dic)+1)
        elif c[k]==a[i] and c[k] != b[j]:
            count = min(calculate(i,j+1,k+1,dic)+1,calculate(i+1,j,k+1,dic))
        else:
            count = min(calculate(i+1,j,k+1,dic),calculate(i,j+1,k+1,dic))

        dic[key_] = count
        return count

    print(calculate(0,0,0,dic))

**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
def
abcdef
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

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

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
import java.util.Scanner;
public class Main
{
    private static int helper(int i,int j,int n1,int n2,int k,String a,String b,String c)
    {
        if(i==n1)
        {
            int count=0;
            for(int idx=j;idx<n2;idx++)
            {
                if(b.charAt(idx)!=c.charAt(k))
                {
                    count++;
                }
                k++;
            }
            return count;
        }

        if(j==n2)
        {
            int count=0;
            for(int idx=i;idx<n1;idx++)
            {
                if(a.charAt(idx)!=c.charAt(k))
                {
                    count++;
                }
                k++;
            }
            return count;
        }

        int min = helper(i+1, j, n1, n2, k+1, a, b, c) + (a.charAt(i) != c.charAt(k) ? 1 : 0);
        min = Math.min(min, helper(i, j+1, n1, n2, k+1, a, b, c) + (b.charAt(j) != c.charAt(k) ? 1 : 0));

        return min;

    }
	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();
		    String c=sc.nextLine();
		    int n1=a.length();
		    int n2=b.length();
		    int ans=helper(0,0,n1,n2,0,a,b,c);
		    System.out.println(ans);
		}
	}
}

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

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

Think step by step.

Response:
```plaintext
1
abc
def
abcdef
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

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

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

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

public class Main {
	static String a,b,c;
	static int[][][] dp;
	static int solve(int ind1, int ind2, int ind3) {
		if(ind3 >= c.length() || (ind1 >= a.length() && ind2 >= b.length())) {
			return (c.length() - (ind3+1));
		}
		if(dp[ind1][ind2][ind3] != -1) return dp[ind1][ind2][ind3];
		int res = Integer.MAX_VALUE;
		if(ind1 < a.length() && a.charAt(ind1) == c.charAt(ind3)) {
			res = Math.min(res, 0+solve(ind1 + 1, ind2, ind3 + 1));
		}
		if(ind2 < b.length() && b.charAt(ind2) == c.charAt(ind3)) {
			res = Math.min(res, 0+solve(ind1, ind2 + 1, ind3 + 1));
		}
		if(ind1 < a.length()) res = Math.min(res, 1+solve(ind1+1, ind2, ind3+1));
		if(ind2 < b.length()) res = Math.min(res, 1+solve(ind1, ind2 + 1, ind3 + 1));
		return dp[ind1][ind2][ind3] = res;
	}
	public static void main(String[] args) {
		FastIO sc = new FastIO();
		FastWriter out = new FastWriter();

		int t= sc.nextInt();
		while(t-- > 0) {
			a = sc.next();
			b = sc.next();
			c = sc.next();
			dp = new int[a.length()+1][b.length()+1][c.length()+1];
			for(int[][] it2 : dp) for(int[] it : it2) Arrays.fill(it,-1);
			int res = solve(0,0,0);
			out.println(res+1);
		}

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

		// #region FastIO and FastWriter
	static class FastIO extends PrintWriter {
		private InputStream stream;
		private byte[] buf = new byte[1 << 16];
		private int curChar, numChars;

		public FastIO() {
			this(System.in, System.out);
		}

		public FastIO(InputStream i, OutputStream o) {
			super(o);
			stream = i;
		}

		public FastIO(String i, String o) throws IOException {
			super(new FileWriter(o));
			stream = new FileInputStream(i);
		}

		private int nextByte() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars == -1)
					return -1; // End of file
			}
			return buf[curChar++];
		}

		public String next() {
			int c;
			do {
				c = nextByte();
			} while (c <= ' ');
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = nextByte();
			} while (c > ' ');
			return res.toString();
		}

		public int nextInt() {
			int c;
			do {
				c = nextByte();
			} while (c <= ' ');
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = nextByte();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res = 10 * res + c - '0';
				c = nextByte();
			} while (c > ' ');
			return res * sgn;
		}

		public long nextLong() {
			int c;
			do {
				c = nextByte();
			} while (c <= ' ');
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = nextByte();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res = 10 * res + c - '0';
				c = nextByte();
			} while (c > ' ');
			return res * sgn;
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public String nextLine() {
			StringBuilder res = new StringBuilder();
			int c = nextByte();

			// If the current character is a new line, read the next character
			if (c == '\n') {
				c = nextByte();
			}

			// Read until the end of the line or the end of the stream
			while (c != -1 && c != '\n') {
				res.appendCodePoint(c);
				c = nextByte();
			}

			return res.toString();
		}

		public char nextChar() {
			int c = nextByte();
			while (c <= ' ')
				c = nextByte();
			return (char) c;
		}

		public float nextFloat() {
			return Float.parseFloat(next());
		}
	}

	static class FastWriter {
		private final StringBuilder sb;

		public FastWriter() {
			this.sb = new StringBuilder();
		}

		public void print(Object object) {
			sb.append(object);
		}

		public void println(Object object) {
			sb.append(object).append("\n");
		}

		public void print(char c) {
			sb.append(c);
		}

		public void println(char c) {
			sb.append(c).append("\n");
		}

		public void print(int i) {
			sb.append(i);
		}

		public void println(int i) {
			sb.append(i).append("\n");
		}

		public void print(long l) {
			sb.append(l);
		}

		public void println(long l) {
			sb.append(l).append("\n");
		}

		public void print(double d) {
			sb.append(d);
		}

		public void println(double d) {
			sb.append(d).append("\n");
		}

		public void close() throws IOException {
			System.out.print(sb);
		}
	}
		// #endregion
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that could cause the algorithm to fail. The buggy code has an issue in the way it calculates the minimum number of changes required. Specifically, it incorrectly adds 1 to the result in some cases where it shouldn't.

Let's break down the problem and identify potential edge cases:

1. **Empty Strings**: While the problem guarantees that the strings are non-empty, we should still consider very short strings to see if the algorithm handles them correctly.
2. **Strings with Identical Characters**: If `a` and `b` have many identical characters, the algorithm might incorrectly count the changes.
3. **Strings with No Matching Characters**: If `a` and `b` have no characters that match `c`, the algorithm should correctly count all characters as changes.
4. **Strings with Partial Matches**: If `a` and `b` have some matching characters but not all, the algorithm should correctly count the minimum changes required.
5. **Long Strings**: While the problem constraints are manageable, we can still use moderately long strings to test the algorithm's performance and correctness.

Given these considerations, let's create a test input that includes a combination of these edge cases:

```plaintext
4
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
```

This test input includes:
1. A simple case with short strings (`a` and `b` with one character each, and `c` with two characters).
2. A case with partial matches (`a` and `b` have some matching characters, and `c` has a mix of characters from both).
3. A case with no matching characters (`a` and `b` have no characters that match `c`).
4. A case with a mix of matching and non-matching characters (`a` and `b` have some matching characters, and `c` has a mix of characters from both).

This test input should help expose any issues in the buggy code.

4
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

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

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
int solve(int i, int j, string a, string b, string c, vector<vector<int>>& dp){
    if(i + j >= c.size()) return 0;
    if(dp[i][j] != 0) return dp[i][j];

    int l = (i < a.size()) ? (((a[i] == c[i+j]) ? 0 : 1) + solve(i+1, j, a, b, c, dp)) : 1e9;
    int r = (j < b.size()) ? ((b[j] == c[i+j] ? 0 : 1) + solve(i, j+1, a, b, c, dp)) : 1e9;

    dp[i][j] = min(l, r);

    return dp[i][j];
}

void solve(){
    int t;
    cin>>t;
    while(t--){
        string a, b, c;
        cin>>a>>b>>c;

        vector<vector<int>>dp(a.length()+1, vector<int>(b.length()+1, 0));

        int ans = solve(0, 0, a, b, c, dp);

        cout<<ans<<"\n";
    }
}

int32_t main(){
    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
ab
ba
abab
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

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

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
# Author: rishabhxchoudhary
# Created: Friday, 27.12.2024 01:20 PM (GMT+5:30)
import sys
sys.setrecursionlimit(2147483647)
MOD = 1000000007
input = sys.stdin.readline

def solve():
    a = input().strip()
    b = input().strip()
    c = input().strip()
    n = len(a)
    m = len(b)
    memo = [[0 for _ in range(m+1)] for _ in range(n+1)]
    def dp(j=0, k=0):
        nonlocal memo
        if memo[j][k]:
            return memo[j][k]
        i = j + k
        if i==len(c): return 0
        if (j<n and c[i]==a[j]) and (k<m and c[i]==b[k]):
            ans = min(dp(j,k+1), dp(j+1,k))
        elif (j<n and c[i]==a[j]):
            if k==m:
                ans = dp(j+1,k)
            else:
                ans = min(1+dp(j,k+1), dp(j+1,k))
        elif (k<m and c[i]==b[k]):
            if j==n:
                ans = dp(j,k+1)
            else:
                ans = min(1+dp(j+1,k), dp(j,k+1))
        else:
            if j<n and k<m:
                ans = 1 + min(dp(j,k+1), dp(j+1,k))
            elif j<n:
                ans = 1 + dp(j+1,k)
            else:
                ans = 1 + dp(j,k+1)
        memo[j][k] = ans
        return ans
    print(dp())

t = int(input())
for _ in range(t):
    solve()

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

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

Think step by step.

Response:
```plaintext
1
abc
def
abcdef
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

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

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
# ======= author: kuanhungchen.com | created: 05/02/23 08:33:06 CST ======= #
from sys            import stdin,  stderr, stdout, setrecursionlimit
from bisect         import bisect_left, bisect_right
from collections    import defaultdict, deque, Counter
from itertools      import accumulate, combinations, permutations, product
from functools      import lru_cache, cmp_to_key, reduce
from heapq          import heapify, heappush, heappop, heappushpop, heapreplace
from pypyjit        import set_param
set_param("max_unroll_recursion=-1")
# setrecursionlimit(300005)
orda  = 97
ordA  = 65
INF   = 1 << 60
MOD   = 1000000007
MOD2  = 998244353
input = lambda: stdin.readline().rstrip("\r\n")
linen = lambda cf_: "[L" + str(cf_.f_back.f_lineno) + "] "
dbg   = lambda *A, **M: stderr.write("\033[91m" + \
        M.get("sep", " ").join(map(str, A)) + M.get("end", "\n") + "\033[0m")
# ============================ START OF MY CODE ============================ #

def solve(_tc):
    a = input()
    b = input()
    c = input()

    @lru_cache(None)
    def dp(i, j):
        ans = 0
        if i == len(a):
            for jj in range(j, len(b)):
                ans += int(c[i + jj] != b[jj])
        elif j == len(b):
            for ii in range(i, len(a)):
                ans += int(c[j + ii] != a[ii])
        else:
            ans = int(c[i + j] != a[i]) + dp(i + 1, j)
            ans = min(ans, int(c[i + j] != b[j]) + dp(i, j + 1))
        return ans

    print(dp(0, 0))

if __name__ == "__main__":
    _tcs = int(input())
    for _tc in range(1, vars().get("_tcs", 1) + 1):
        dbg("=== Case {} ===".format(str(_tc).rjust(2)))
        solve(_tc)

**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
abcdef
abc
def
abdcfe
abc
def
fedcba
```

3
abc
def
abcdef
abc
def
abdcfe
abc
def
fedcba


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

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

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
#include <stdio.h>
#include <string.h>

#define MIN(x, y) ((x)<=(y)?(x):(y))

const int N = 1e3+5, inf = N*3;

int dp[N][N];

int solver(const char* a, const char* b, const char* c){
    int n = strlen(a), m = strlen(b), k = strlen(c);
    for(int i=0; i<=k; ++i)
        for(int j=0; j<=n; ++j)
            dp[i][j] = inf;
    dp[0][0] = 0;
    for(int i=1; i<=k; ++i){
        for(int j=0; j<=MIN(i,n); ++j){
            if(i-j>m) continue;
            int val1 = (i-j-1>=0) ? (dp[i-1][j]+(b[i-j-1]!=c[i-1])) : inf;
            int val2 = (j-1>=0) ? dp[i-1][j-1]+(a[j-1]!=c[i-1]) : inf;
            dp[i][j] = MIN(val1, val2);
        }
    }
    return dp[k][n];
}

int main(void){
    int T;
    char a[N], b[N], c[N];
    scanf("%d", &T);
    while(T--){
        scanf("%s%s%s", a, b, c);
        printf("%d\n", solver(a, b, c));
    }
    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
def
abcdegf
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

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

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

Buggy Code:
import java.util.*;
public class A {
    private static int isInterleaveHelper(String s1, String s2, String s3, int i, int j, Map<String, Integer> memo) {
        // Base case: if we have reached the end of s3, we successfully interleaved
        if (i==s1.length() && j == s2.length()) return 0;

        // Create a unique key for the current state
        String key = i + "," + j;
        if (memo.containsKey(key)) return memo.get(key);
        int takeFromS1 = 10000000 , takeFromS2 = 10000000;

        if(i<s1.length())
        takeFromS1 = isInterleaveHelper(s1, s2, s3, i + 1, j,  memo)  + (s1.charAt(i)==s3.charAt(i+j)? 0:1) ;

        if(j<s2.length())
        takeFromS2 = isInterleaveHelper(s1, s2, s3, i, j + 1,  memo) +  (s2.charAt(j)==s3.charAt(i+j)? 0:1);

        int result = Math.min(takeFromS1,takeFromS2);
        memo.put(key, result);
        return result;
    }
    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 = sc.next();

            int result = isInterleaveHelper(a, b, c, 0, 0, new HashMap<>());
            System.out.println(result);

        }
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided code, we need to consider edge cases that might cause the algorithm to fail. The bug in the provided code is that it does not correctly handle cases where the characters in `c` do not match the characters in `a` or `b` when they should. This can happen when the characters in `c` are not in the correct order or when the characters in `c` do not match the characters in `a` or `b` at all.

### Step-by-Step Analysis:
1. **Equal Lengths of `a` and `b`:** The code should handle cases where `a` and `b` are of equal length.
2. **Unequal Lengths of `a` and `b`:** The code should handle cases where `a` and `b` are of different lengths.
3. **Character Mismatch:** The code should handle cases where characters in `c` do not match the characters in `a` or `b` at all.
4. **Order of Characters:** The code should handle cases where the order of characters in `c` does not match the order of characters in `a` or `b`.

### Test Case 1: Mismatched Characters
- `a = "abc"`
- `b = "def"`
- `c = "adbecf"`
  - Here, `c` is a valid interleaving of `a` and `b` but with some characters swapped. The expected output is 3, as all characters in `c` need to be changed to match `a` and `b`.

### Test Case 2: Different Lengths
- `a = "abc"`
- `b = "defg"`
- `c = "adbecfg"`
  - Here, `c` is a valid interleaving of `a` and `b` but with some characters swapped. The expected output is 4, as four characters in `c` need to be changed to match `a` and `b`.

### Test Case 3: Characters in `c` do not match `a` or `b`
- `a = "abc"`
- `b = "def"`
- `c = "xyz"`
  - Here, `c` does not match any characters in `a` or `b`. The expected output is 6, as all characters in `c` need to be changed to match `a` and `b`.

### Test Case 4: Order of Characters
- `a = "abc"`
- `b = "def"`
- `c = "fedcba"`
  - Here, `c` is a reversed interleaving of `a` and `b`. The expected output is 6, as all characters in `c` need to be changed to match `a` and `b`.

### Test Case 5: Single Character Strings
- `a = "a"`
- `b = "b"`
- `c = "b"`
  - Here, `c` is a valid interleaving of `a` and `b` but with one character swapped. The expected output is 1, as one character in `c` needs to be changed to match `a` and `b`.

### Test Case 6: Empty Strings
- `a = ""`
- `b = ""`
- `c = "abc"`
  - Here, `c` does not match any characters in `a` or `b`. The expected output is 3, as all characters in `c` need to be changed to match `a` and `b`.

### Final Test Input
```plaintext
6
abc
def
adbecf
abc
defg
adbecfg
abc
def
xyz
abc
def
fedcba
a
b
b
""
""
abc
```

6
abc
def
adbecf
abc
defg
adbecfg
abc
def
xyz
abc
def
fedcba
a
b
b
""
""
abc


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Three Strings
time_limit_ms: 2500
memory_limit_mb: 256
Description: You are given three strings: $$$a$$$, $$$b$$$, and $$$c$$$, consisting of lowercase Latin letters. The string $$$c$$$ was obtained in the following way:

1. At each step, either string $$$a$$$ or string $$$b$$$ was randomly chosen, and the first character of the chosen string was removed from it and appended to the end of string $$$c$$$, until one of the strings ran out. After that, the remaining characters of the non-empty string were added to the end of $$$c$$$.
2. Then, a certain number of characters in string $$$c$$$ were randomly changed.

For example, from the strings $$$a=\color{red}{\text{abra}}$$$ and $$$b=\color{blue}{\text{cada}}$$$, without character replacements, the strings $$$\color{blue}{\text{ca}}\color{red}{\text{ab}}\color{blue}{\text{d}}\color{red}{\text{ra}}\color{blue}{\text{a}}$$$, $$$\color{red}{\text{abra}}\color{blue}{\text{cada}}$$$, $$$\color{red}{\text{a}}\color{blue}{\text{cada}}\color{red}{\text{bra}}$$$ could be obtained.

Find the minimum number of characters that could have been changed in string $$$c$$$.

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

The first line of each test case contains one string of lowercase Latin letters $$$a$$$ ($$$1 \leq |a| \leq 10^3$$$) — the first string, where $$$|a|$$$ denotes the length of string $$$a$$$.

The second line of each test case contains one string of lowercase Latin letters $$$b$$$ ($$$1 \leq |b| \leq 10^3$$$) — the second string, where $$$|b|$$$ denotes the length of string $$$b$$$.

The third line of each test case contains one string of lowercase Latin letters $$$c$$$ ($$$|c| = |a| + |b|$$$) — the third string.

It is guaranteed that the sum of $$$|a|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$. Also, the sum of $$$|b|$$$ across all test cases does not exceed $$$2 \cdot 10^3$$$.

Output: For each test case, output a single integer — the minimum number of characters that could have been changed in string $$$c$$$.

Examples:
input:
7
a
b
cb
ab
cd
acbd
ab
ba
aabb
xxx
yyy
xyxyxy
a
bcd
decf
codes
horse
codeforces
egg
annie
egaegaeg
output:
1
0
2
0
3
2
3

Note:

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

public class Main {
	static long cr[][]=new long[1001][1001];
	//static double EPS = 1e-7;
   static long mod=998244353;
   static long val=0;
   static boolean flag=true;

public static void main(String[] args)
{
	FScanner sc = new FScanner();
	//Arrays.fill(prime, true);
	//sieve();
	//ncr();
	int t=sc.nextInt();
	StringBuilder sb = new StringBuilder();

     while(t-->0)
     {
    	char a[]=sc.nextLine().toCharArray();
    	char b[]=sc.nextLine().toCharArray();
    	char c[]=sc.nextLine().toCharArray();
    	 HashMap<String, Integer> hMap=new HashMap<>();
    	 sb.append(Check(a,b,c,a.length-1,b.length-1,c.length-1,hMap));

	    sb.append("\n");
     }

      System.out.println(sb.toString());
}
public static int Check(char a[],char b[],char c[],int i,int j,int k,HashMap<String, Integer> hMap)
{
	if(i<0 && j<0 && k<0) {
		return 0;
	}
	if(k<0) {
		return 100000;
	}
	if(hMap.containsKey(i+" "+j+" "+k)) {
		return hMap.get(i+" "+j+" "+k);
	}
	 int res=100000;
	if(i>=0 && k>=0 && a[i]==c[k]) {
		res= Check(a, b, c, i-1, j, k-1,hMap);
	}
	if(j>=0 && k>=0 && b[j]==c[k]) {
		res=Math.min( res,Check(a, b, c, i, j-1, k-1,hMap));
	}

	res = Math.min(res,1+Math.min(Check(a, b, c, i-1, j, k-1,hMap), Check(a, b, c, i, j-1, k-1,hMap)));
	hMap.put(i+" "+j+" "+k, res);
	return res;
}

public static long powerLL(long x, long n)
{
    long result = 1;
    while (n > 0)
    {
        if (n % 2 == 1)
        {
            result = result * x % mod;
        }
        n = n / 2;
        x = x * x % mod;
    }
    return result;
}

public static long gcd(long a,long b)
{

     return b==0 ? a:gcd(b,a%b);

}

 /* public static void sieve()
{   prime[0]=prime[1]=false;
int n=1000000;
	for(int p = 2; p*p <=n; p++)
    {
        if(prime[p] == true)
        {

            for(int i = p*p; i <= n; i += p)
                prime[i] = false;
        }
    }
   */

public static void ncr()
{
	cr[0][0]=1;
	for(int i=1;i<=1000;i++)
	{
		cr[i][0]=1;
		for(int j=1;j<i;j++)
		{
			cr[i][j]=(cr[i-1][j-1]+cr[i-1][j])%mod;
		}
		cr[i][i]=1;
	}
}

}

  class pair
  //implements Comparable<pair>
  {
	 long a;long b;
	 pair(long a,long b)
	{
		this.b=b;
		this.a=a;
	}

  }

 class FScanner {
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer sb = new StringTokenizer("");

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

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

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

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

	float nextFloat(){
	return Float.parseFloat(next());
	}

	double nextDouble(){
	return Double.parseDouble(next());
	}
}

**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 and scenarios that can highlight the issues in the algorithm. The main issues in the buggy code include:

1. **Handling of the base cases**: The code might not correctly handle cases where one of the strings runs out of characters.
2. **Memoization**: The memoization might not be implemented correctly, leading to redundant calculations or incorrect results.
3. **Character matching**: The code might not correctly handle the minimum number of character changes required.

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

1. **Strings of different lengths** to ensure the code handles the termination conditions correctly.
2. **Strings with repeated characters** to test the memoization and character matching.
3. **Strings that require character changes** to verify the correctness of the minimum changes calculation.

Here is a small-scale test input that meets these criteria:

```plaintext
3
a
b
cb
ab
ba
aabb
xxx
yyy
xyxyxy
```

### Explanation of the Test Input:
1. **Test Case 1**:
   - \( a = "a" \)
   - \( b = "b" \)
   - \( c = "cb" \)
   - Expected Output: 1 (The string "cb" can be obtained by changing the 'a' in "a" to 'c' and then appending 'b' from "b".)

2. **Test Case 2**:
   - \( a = "ab" \)
   - \( b = "ba" \)
   - \( c = "aabb" \)
   - Expected Output: 2 (The string "aabb" can be obtained by changing the 'b' in "b" to 'a' and the 'a' in "a" to 'b'.)

3. **Test Case 3**:
   - \( a = "xxx" \)
   - \( b = "yyy" \)
   - \( c = "xyxyxy" \)
   - Expected Output: 0 (The string "xyxyxy" can be obtained by alternating characters from "xxx" and "yyy" without any changes.)

This test input should expose the bugs in the given code by challenging its handling of different string lengths, repeated characters, and the minimum number of character changes required.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
#include <iostream>
#include <vector>
#define long long int
using namespace std;

signed main(){
    int n_test;
    cin >> n_test;
    while(n_test--){
        int n;
        cin >> n;
        int remainder = n%4;
        if (remainder > 1 && remainder <3){
            cout<<"Sakurako\n";
        }
        else{
            cout<<"Kosuke\n";
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
2
4
5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
#include <iostream>
#include <vector>
using namespace std;
#define ll long long
int n;
int main()
{
int t;cin>>t;
while(t--)
{
    cin>>n;int i=0,z=1;
    while(i<=n&&i>=(-1*n))
    {
        if(z==1)
        {
            i+=(i*2+1);z=0;
        }
        else {i+=(i*2+1);z=1;}
    }
    if(z==0)cout<<"Sakurako\n";
    else cout<<"Kosuke\n";z=1;
}
}

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

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

Think step by step.

Response:
```plaintext
3
1
2
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
// Problem: A. Sakurako and Kosuke
// Contest: Codeforces - Codeforces Round 981 (Div. 3)
// URL: https://codeforces.com/contest/2033/problem/0
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#include <iostream>
using namespace std;
#define utkarsh 1
#define JAI ios_base::sync_with_stdio(false);
#define SHREE cin.tie(NULL);
#define RAM cout.tie(NULL);
using ll= long long;
using sg=string;
using lld= long double;
using ull= unsigned long long;
const lld pi= 3.141592653589793238;
const ll mod=1e9+7;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<vll> vvll;
typedef vector<string> vs;
typedef unordered_map<ll,ll> umll;
typedef map<ll,ll> mll;
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define fl(i,n) for(int i=0;i<n;i++)
#define rl(i,m,n) for(int i=n;i>=m;i--)
#define py cout<<"YES\n";
#define pm cout<<"-1\n";
#define pn cout<<"NO\n";
#define tl(n)  ll n;cin>>n;
#define ts(n)  sg n;cin>>n;
#define vr(v) v.begin(),v.end()
#define rv(v) v.end(),v.begin()
#define memz(a) memset(a, 0, sizeof(a))
#define memn(a) memset(a, -1, sizeof(a))
#define p(x) cout<<x<<endl;
#define flp(i,a,b) for(int i = a; i <= b; i++)

/*------------------------------Graph Moves----------------------------*/
// const int fx[]={+1,-1,+0,+0};
// const int fy[]={+0,+0,+1,-1};
// const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
// const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
// const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
// const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*---------------------------------------------------------------------*/

//Debug
#ifdef utkarsh
#define debug(x) cerr << #x<<" "; cerr<<x<<" "; cerr << endl;
#else
#define debug(x);
#endif

// Operator overloads
template<typename T1, typename T2> // cin >> pair<T1, T2>
istream& operator>>(istream &istream, pair<T1, T2> &p) { return (istream >> p.first >> p.second); }
template<typename T> // cin >> vector<T>
istream& operator>>(istream &istream, vector<T> &v){for (auto &it : v)cin >> it;return istream;}
template<typename T1, typename T2> // cout << pair<T1, T2>
ostream& operator<<(ostream &ostream, const pair<T1, T2> &p) { return (ostream << p.first << " " << p.second); }
template<typename T> // cout << vector<T>
ostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << " "; return ostream; }

// Utility functions
template <typename T>
void print(T &&t)  { cout << t << "\n"; }
void printarr(ll arr[], ll n){fl(i,n) cout << arr[i] << " ";cout << "\n";}
template<typename T>
void printvec(vector<T>v){ll n=v.size();fl(i,n)cout<<v[i]<<" ";cout<<"\n";}
template<typename T>
ll sumvec(vector<T>v){ll n=v.size();ll s=0;fl(i,n)s+=v[i];return s;}

void square(){
    //write code here
    tl(n);
    ll c=0;
    flp(i,1,100){
    	ll sp=2*i-1;
    	if(i%2==1){
    		c-=sp;
    	}else{
    		c+=sp;
    	}
    	if(abs(c)>n){
    		if(i%2){
    			cout<<"Sakurako"<<endl;
    		}else{
    			cout<<"Kosuke"<<endl;
    		}
    		return;
    	}
    }

 }

int main()
{
    JAI SHREE RAM
    int n = 1;
    cin>>n;
    while(n--){
       square();
    }
}

**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 the following:

1. **Edge Cases**: The boundary values for `n` (1 and 100) should be tested.
2. **Odd and Even Values**: Since the game alternates between Sakurako and Kosuke, and the result can depend on whether the number of moves is odd or even, we should test both.
3. **Values that Cause the Game to End Early**: Values that cause the game to end quickly, such as small values of `n`, can help identify issues in the logic.
4. **Values that Cause the Game to Last Longer**: Values that cause the game to last longer, such as larger values of `n`, can help identify issues in the logic that handle more moves.

Given these considerations, a small-scale test input that covers these scenarios would be:

```plaintext
5
1
2
3
4
100
```

This test input includes:
- The smallest value (1), which should end the game quickly.
- The next few small values (2, 3, 4) to test both odd and even values and their impact on the game.
- The largest value (100) to test a scenario where the game lasts longer.

This input should help expose any bugs in the logic, especially in handling the game's termination condition and the alternation between Sakurako and Kosuke.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
n = int(input())

for i in range(n):
    m = int(input())
    if (i + 1) % 2 == 0:
        print('Sakurako')
    else:
        print('Kosuke')

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

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

Think step by step.

Response:
```plaintext
3
1
2
3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
x = 0
flag = 0
name = "Sakurako"
t = int(input())
for k in range(t):
    n = int(input())
    #print(n)
    #for j in range (1,n-1):
        #l = 2 * j -1
        #if (flag%2 ==0):
            #x-=l
        #else:
            #x+=l
    #print(x)

    if (flag%2 ==0): #четный
        name = "Kosuke"
    else:
        name = "Sakurako"
    flag+=1
    print(name)

**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
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
def solve(n):
    if(n==1):
        print("Kosuke")
        return
    s=0
    i=0
    l=0
    while(abs(s)<=n):
        if(i%2==0):
            l=(2*l-1)
            s=s-abs(l)

        else:
            l=(2*l-1)
            s=s+abs(l)
        # print(s)
        i+=1
    if(i%2==1):
        print("Kosuke")
    else:
        print("Sakurako")

for _ in range(int(input())):
    n=int(input())
    solve(n)
    # print(ans)

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

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

Think step by step.

Response:
```plaintext
4
1
2
3
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
import java.util.Scanner;
public class SakurakoAndKosuke {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int p,x=0;
        int t= sc.nextInt();
        int n[]=new int[t+1];
        for(int i=1;i<=t;i++){
            n[i]=sc.nextInt();
        }
        for(int j=0;j<t;j++){
            if(j%2==0){
                System.out.println("Kosuke");
            }
            else{
                System.out.println("Sakurako");
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1
2
3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
import java.util.Scanner;

public class Game{
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            if (n == 1) {
                System.out.println("Kosuke");
                continue;
            }
            int rem = n % 2;
            n = n / 2 + rem;
            int s = n % 2;
            s = s / 2 + s;

            if (s % 2 == 0) {
                System.out.println("Kosuke");
            } else {
                System.out.println("Sakurako");
            }
        }
        sc.close();
    }

}

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

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

Think step by step.

Response:
```plaintext
3
1
2
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
// Code by "sword".
// University of Science, VNU-HCM.
//------------------------------------------------------//

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;

#define int long long

#define pii pair<int,int>
#define pll pair<ll, ll>

#define mp(a,b) make_pair(a,b)
#define fi first
#define se second

#define vi vector<int>
#define vll vector<ll>
#define vpii vector<pii>
#define vs vector<string>

#define For(i,a,n) for (int i = a; i <= n; i++)
#define ForD(i,a,n) for (int i = a; i >= n; i--)
#define trav(x,a) for (auto x: a)

#define pb push_back
#define eb emplace_back

#define num1(x) __builtin_popcount(x) // count number of 1 bit
#define num1ll(x) __builtin_popcountll(x)
#define clz(x) __builtin_clz(x) // count leading zero
#define clzll(x) __builtin_clzll(x)
#define ctz(x) __builtin_ctz(x) // count trailing zero
#define ctzll(x) __builtin_ctzll(x)

#define all(x) (x).begin()+1, (x).end()
#define ALL(x) (x).begin(), (x).end()

#define sfmax(a,b) (a) = max((a), (b))
#define sfmin(a,b) (a) = min((a), (b))

#define endl "\n"
#define MULTI_TEST 1

int bound2(int x) {
    // Compute the smallest power of 2 that is not smaller than x.
    // Example: x = 5 -> return 8, x = 1 -> return 1
    return 1 << (32 - __builtin_clz(x-1));
}

void yes() {cout << "YES\n"; return;}
void no() {cout << "NO\n"; return;}

const int N = 2e5 + 5;
const int inf = INT_MAX;
const long long linf = 1e18;
const int mod = 1e9 + 7;

void solve() {
    int n; cin >> n;
    int t = 1;
    int pos = -1;
    if (n == 1) {
        cout << "Kosuke" << endl;
        return;
    }
    else {
        while (pos <= n and pos >= -n) {
            t += 1;
            pos = 2*t - 1;
            if (t % 2 == 1) pos = -pos;
        }
        cout << (t % 2 != 0 ? "Kosuke" : "Sakurako") << endl;
    }

}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    if (MULTI_TEST) {
        int tc; cin >> tc;
        while (tc--) solve();
    }
    else 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
4
1
2
3
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
  int t; cin >> t;
  while(t--){
    int n; cin >> n;
    int ans = 0;
    for(int i = 0; i < 100; i++){
      if(i & 1){
        ans += 2*(i+1)-1;
      }else{
        ans -= 2*(i+1)-1;
      }
      if(abs(ans) > n){
        if(i % 2 == 0){
          cout << "Sakurako";
        }else{
          cout << "Kosuke";
        }
        break;
      }
    }
    cout << endl;
  }
}

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

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

Think step by step.

Response:
```plaintext
1
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
t = int(input())
for i in range(t):
    n = int(input())
    a = 1
    i = 1
    ans = 0
    while a < n:
        if i % 2 != 0:
            ans -= a
        else:
            ans += a
        a = i * 2 - 1
        i += 1
    if ans == 0:
        print("Kosuke")
    else:
        print("Sakurako")

**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 can maximize the failure probability. The buggy code has a logical issue in how it calculates the position of the dot and determines the winner. Specifically, it incorrectly updates the value of `a` and `i` within the loop.

Here are the steps to create a challenging test input:

1. **Edge Case with Small `n`**:
   - When `n` is small, the game can end quickly. This can help identify issues in the initial stages of the loop.

2. **Edge Case with Large `n`**:
   - When `n` is large, the game will have more turns, which can help identify issues in the loop's logic and the final position calculation.

3. **Edge Case with `n` that Causes Specific Position**:
   - Choose an `n` that causes the position to be exactly at a boundary condition, such as `x = n` or `x = -n`.

4. **Multiple Test Cases**:
   - Include multiple test cases to cover different scenarios and ensure the code handles all cases correctly.

Given these considerations, a challenging test input would be:

```plaintext
4
1
2
3
4
```

This test input includes small values of `n` that will cause the game to end quickly and different boundary conditions. It will help expose any issues in the loop logic and the final position calculation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
t = int(input());
def ehe(x):
    if (x%2==0): return "Kosuke"
    else: return "Sakurako"
for i in range (t):
    n = int(input()); x = 0;
    for i in range (1, 100):
        x += ((-1)**i)*(2*i-1);
        if abs(x) > n:
            print (ehe(i));
            break;

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the code might fail. The buggy code has a logic error in determining the winner based on the turn number. Specifically, it incorrectly returns "Kosuke" for even turns and "Sakurako" for odd turns.

Let's consider the following steps to create a test input:

1. **Edge Cases for Small Values of \( n \):**
   - For \( n = 1 \), the game ends after the first move.
   - For \( n = 2 \), the game ends after the second move.
   - For \( n = 3 \), the game ends after the third move.

2. **Cases Where the Absolute Value of \( x \) Reaches \( n \) Exactly:**
   - For \( n = 5 \), the game ends after the fourth move.
   - For \( n = 7 \), the game ends after the fifth move.

3. **Cases Where the Absolute Value of \( x \) Exceeds \( n \) Just After a Move:**
   - For \( n = 4 \), the game ends after the third move.
   - For \( n = 6 \), the game ends after the fourth move.

4. **Larger Values of \( n \) to Test the General Case:**
   - For \( n = 9 \), the game ends after the fifth move.
   - For \( n = 10 \), the game ends after the sixth move.

Based on these considerations, a small-scale test input that includes a mix of these cases is:

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

This test input includes:
- \( n = 1 \) (Sakurako wins)
- \( n = 2 \) (Kosuke wins)
- \( n = 3 \) (Sakurako wins)
- \( n = 4 \) (Kosuke wins)
- \( n = 5 \) (Sakurako wins)
- \( n = 6 \) (Kosuke wins)

This will help expose the bug in the faulty logic of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

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

public class Main {

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int x = 0;
            int count = 0;
            while (Math.abs(x) <= n) {
                x += 2 * x - 1;
                count++;
            }
            if (count % 2 == 0) {
                System.out.println("Kosuke");
            }
            else {
                System.out.println("Sakurako");
            }
        }
        sc.close();
    }

}

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

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

Think step by step.

Response:
```plaintext
4
1
2
3
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
import java.util.Scanner;

public class step {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        int count = 0;
        for (int i = 1; i <= t; i++) {
           int n = scan.nextInt();
           int every = 2 * i - 1;
           if(every <= n + 2){
               count++;
           }
           if(count % 2 == 0){
               System.out.println("Sakurako");
           }else{
               System.out.println("Kosuke");
           }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1
2
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

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

using namespace std;

//#define int long long
#define pii pair<int, int>
#define fi first
#define se second
#define rep(i, n) for(int i = 0; i < (n); i++)

typedef long long ll;
typedef long double ld;

const int MAXN = 1e6 + 7;
const int INF = 1e9 + 7;
const int MAX = 3e3 + 7;
const int MOD = 1e9 + 7;
const int base = 1<<20;
const ld EPS = 1e-9;

void solve(){

	int n;
	cin >> n;

	int ok = 1;
	int pos = 0;

	for(int i = 1; i <= 100; i++){

		if(ok) pos -= (2 * i - 1);
		else pos += (2 * i - 1);

		if(abs(pos) > n){

			if(!ok) cout << "Kosuke\n";
			else cout << "Sakurako\n";

			return;
		}
		ok ^= 1;
	}

}

bool multi = 1;

signed main(){

	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);

	int t = 1;
	if(multi) 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
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako and Kosuke
time_limit_ms: 1000
memory_limit_mb: 256
Description: Sakurako and Kosuke decided to play some games with a dot on a coordinate line. The dot is currently located in position $x=0$. They will be taking turns, and Sakurako will be the one to start.

On the $i$-th move, the current player will move the dot in some direction by $2\cdot i-1$ units. Sakurako will always be moving the dot in the negative direction, whereas Kosuke will always move it in the positive direction.

In other words, the following will happen:

  1. Sakurako will change the position of the dot by $-1$, $x = -1$ now    2. Kosuke will change the position of the dot by $3$, $x = 2$ now    3. Sakurako will change the position of the dot by $-5$, $x = -3$ now    4. $\cdots$

They will keep on playing while the absolute value of the coordinate of the dot does not exceed $n$. More formally, the game continues while $-n\le x\le n$. It can be proven that the game will always end.

Your task is to determine who will be the one who makes the last turn.

Input: The first line contains one integer $t$ ($1\le t\le 100$) — the number of games that Sakurako and Kosuke played.

Each game is described by one number $n$ ($1 \le n\le 100$) — the number that defines the condition when the game ends.

Output: For each of the $t$ games, output a line with the result of that game. If Sakurako makes the last turn, output "Sakurako" (without quotes); else output "Kosuke".

Examples:
input:
4
1
6
3
98
output:
Kosuke
Sakurako
Kosuke
Sakurako

Note:

Buggy Code:
#include<bits/stdc++.h>
#define ll long long int
#define nl '\n'
#define pi pair<int, int>
#define fastIO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
using namespace std;

void solve()
{
    int x=0, mov= 1, n;
    cin>>n;
    bool sakru= true;
    while (true)
    {

        if (sakru)  x -=mov;
        else x +=mov;

        if (x>n || x< -n)
        {
            if (sakru) cout<<"Sakurako"<<nl;
            else cout<<"Kosuke"<<nl;
            break;

        }
        sakru= !sakru;
        x+=2;
    }

}
int main()
{
    fastIO;

    int test;
    cin>>test;
    while (test--)  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 create a challenging test input that exposes the bug in the provided code, let's analyze the code and identify the potential issues:

1. **Incorrect Movement Calculation**: The code incorrectly increments `x` by 2 after each move, which is not part of the game rules. The correct movement should be `2 * i - 1` for each turn.
2. **Boundary Condition**: The code checks if `x` exceeds `n` or `-n` after the move, but it should check before the move to determine if the game should end.

Given these issues, we need a test case that will make the game end close to the boundary, forcing the code to incorrectly handle the movement and boundary conditions.

Let's consider the following test case:

- **Test Case 1**: `n = 1`
  - The game will end on the first move, and Kosuke will make the last move.
- **Test Case 2**: `n = 3`
  - The game will end on the third move, and Sakurako will make the last move.
- **Test Case 3**: `n = 5`
  - The game will end on the fifth move, and Kosuke will make the last move.
- **Test Case 4**: `n = 7`
  - The game will end on the seventh move, and Sakurako will make the last move.

These test cases will help expose the issues with the incorrect movement and boundary conditions.

Here is the test input:

```plaintext
4
1
3
5
7
```

This input will challenge the code by testing the boundary conditions and the movement logic, ensuring that the bug is exposed.