**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Gerrymandering
time_limit_ms: 2000
memory_limit_mb: 256
Description: We all steal a little bit. But I have only one hand, while my adversaries have two.

Álvaro Obregón

Álvaro and José are the only candidates running for the presidency of Tepito, a rectangular grid of $2$ rows and $n$ columns, where each cell represents a house. It is guaranteed that $n$ is a multiple of $3$.

Under the voting system of Tepito, the grid will be split into districts, which consist of any $3$ houses that are connected$^{\text{∗}}$. Each house will belong to exactly one district.

Each district will cast a single vote. The district will vote for Álvaro or José respectively if at least $2$ houses in that district select them. Therefore, a total of $\frac{2n}{3}$ votes will be cast.

As Álvaro is the current president, he knows exactly which candidate each house will select. If Álvaro divides the houses into districts optimally, determine the maximum number of votes he can get.

$^{\text{∗}}$A set of cells is connected if there is a path between any $2$ cells that requires moving only up, down, left and right through cells in the set.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains one integer $n$ ($3 \le n \le 10^5$; $n$ is a multiple of $3$) — the number of columns of Tepito.

The following two lines each contain a string of length $n$. The $i$-th line contains the string $s_i$, consisting of the characters $\texttt{A}$ and $\texttt{J}$. If $s_{i,j}=\texttt{A}$, the house in the $i$-th row and $j$-th column will select Álvaro. Otherwise if $s_{i,j}=\texttt{J}$, the house will select José.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single integer — the maximum number of districts Álvaro can win by optimally dividing the houses into districts.

Examples:
input:
4
3
AAA
AJJ
6
JAJAJJ
JJAJAJ
6
AJJJAJ
AJJAAA
9
AJJJJAJAJ
JAAJJJJJA
output:
2
2
3
2

Note: The image below showcases the optimal arrangement of districts Álvaro can use for each test case in the example.

![](CDN_BASE_URL/5f1f1034d2d521a09ea4a9a366636fd8)

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

public class gerrymander {
	public static void main(String[] args) throws IOException {
		BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);

		StringTokenizer st = new StringTokenizer(r.readLine());
		int b = Integer.parseInt(st.nextToken());
		for(int z = 0; z<b; z++){
			st = new StringTokenizer(r.readLine());
			int a = Integer.parseInt(st.nextToken());
			int[][] houses = new int[a][2];
			for(int i = 0; i<2; i++){
			 String temp = r.readLine();
			 for(int j = 0; j<a; j++){
			   if(temp.charAt(j) == 'A'){
			     houses[j][i] = 1;
			   }
			   else{
			     houses[j][i] = -1;
			   }
			 }

			}
			int total = 0;
			for(int i = 0; i<a; i+=3){
			   int c = 0;
			   int max = 0;
			   int s1 = houses[i][0]+houses[i+1][0]+houses[i+2][0];
			   int s2 = houses[i][1]+houses[i+1][1]+houses[i+2][1];
			   if(s1>0){
			     c++;
			   }
			   if(s2>0){
			     c++;
			   }
			   if(c>max){
			     max = c;
			   }
			   c = 0;
			   s1 = houses[i][0]+houses[i+1][0]+houses[i][1];
			   s2 = houses[i+1][1]+houses[i+2][0]+houses[i+2][1];
			   if(s1>0){
			     c++;
			   }
			   if(s2>0){
			     c++;
			   }
			   c=0;
			   if(c>max){
			     max = c;
			   }
			   s1 = houses[i+2][1]+houses[i+2][0]+houses[i+1][0];
			   s2 = houses[i][0]+houses[i][1]+houses[i+1][1];
			   if(s1>0){
			     c++;
			   }
			   if(s2>0){
			     c++;
			   }
			   if(c>max){
			     max = c;
			   }
			   total+=max;
			}
			System.out.println(total);

		}

		}
	}

**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
9
AJJJJAJAJ
JAAJJJJJA
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
namespace io{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    inline int read(){
        int x=0; bool flag=true; char c=getchar();
        while(!isdigit(c)){ if(c=='-') flag=false; c=getchar(); }
        while(isdigit(c)){ x=(x<<1)+(x<<2)+(c^48); c=getchar(); }
        return flag?x:~(x-1);
    }
}
using namespace io;
const int N=16;
int T,n,m,a[N][N];
int f[N][1<<N][N];
int g[N][1<<N][N];
int w[N][N],res;
int main(){
    T=read();
    while(T--){
        n=read(),m=read();
        for(int i=0;i<n;i++)
            a[i][m]=0;
        for(int i=0;i<m;i++)
            a[n][i]=0;
        a[n][m]=0,res=1e9;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                a[i][j]=read();
                a[i][m]^=a[i][j];
                a[n][j]^=a[i][j];
                a[n][m]^=a[i][j];
            }
        }
        for(int x=0;x<=n;x++){
            for(int i=0;i<=m-1;i++){
                for(int j=i+1;j<=m;j++){
                    int sum=0;
                    for(int k=0;k<=n;k++){
                        if(k==x) continue;
                        sum+=abs(a[k][i]-a[k][j]);
                    }
                    w[i][j]=w[j][i]=sum;
                }
            }
            for(int i=1;i<1<<m+1;i++)
                for(int j=0;j<=m;j++)
                    f[x][i][j]=1e9;
            for(int i=0;i<=m;i++)
                f[x][1<<i][i]=0;
            for(int i=1;i<1<<m+1;i++){
                for(int j=0;j<=m;j++){
                    if(i>>j&1){
                        for(int k=0;k<=m;k++){
                            if(i>>k&1&&k!=j){
                                f[x][i][j]=min(f[x][i][j],f[x][i^1<<j][k]+w[k][j]);
                            }
                        }
                    }
                }
            }
        }
        for(int x=0;x<=m;x++){
            for(int i=0;i<=n-1;i++){
                for(int j=i+1;j<=n;j++){
                    int sum=0;
                    for(int k=0;k<=m;k++){
                        if(k==x) continue;
                        sum+=abs(a[i][k]-a[j][k]);
                    }
                    w[i][j]=w[j][i]=sum;
                }
            }
            for(int i=1;i<1<<n+1;i++)
                for(int j=0;j<=n;j++)
                    g[x][i][j]=1e9;
            for(int i=0;i<=n;i++)
                g[x][1<<i][i]=0;
            for(int i=1;i<1<<n+1;i++){
                for(int j=0;j<=n;j++){
                    if(i>>j&1){
                        for(int k=0;k<=n;k++){
                            if(i>>k&1&&k!=j){
                                g[x][i][j]=min(g[x][i][j],g[x][i^1<<j][k]+w[k][j]);
                            }
                        }
                    }
                }
            }
        }
        for(int x=0;x<=n;x++){
            for(int y=0;y<=m;y++){
                for(int i=0;i<=n;i++){
                    if(i==x) continue;
                    for(int j=0;j<=m;j++){
                        if(j==y) continue;
                        res=min(res,
                        f[x][(1<<m+1)-1^1<<y][j]+
                        g[y][(1<<n+1)-1^1<<x][i]);
                    }
                }
            }
        }
        printf("%d\n",res);
    }
    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
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

Buggy Code:
//author: towrist

#include <iostream>
#include <iomanip>
#include <vector>
#include <functional>
#include <cmath>
#include <algorithm>
#include <set>
#include <utility>
#include <queue>
#include <map>
#include <assert.h>
#include <stack>
#include <climits>
#include <string>
#include <ctime>
#include <chrono>
#include <random>
#define int long long
using namespace std;

const int INF=1e18;

void solve()
{
	int n, m;
	cin>>n>>m;
	vector<vector<int>> a(n+1, vector<int>(m+1, 0));

	for (int i=0; i<n; i++) for (int j=0; j<m; j++) cin>>a[i][j];

	for (int i=0; i<n; i++)
	{
		int c=0;
		for (int j=0; j<m; j++) c^=a[i][j];
		a[i][m]=c;
	}

	for (int j=0; j<m; j++)
	{
		int c=0;
		for (int i=0; i<n; i++) c^=a[i][j];
		a[n][j]=c;
	}

	{
		int X=0;
		for (int i=0; i<n; i++) for (int j=0; j<m; j++) X^=a[i][j];
		a[n][m]=X;
	}

	vector<vector<int>> ans(n+1, vector<int>(m+1, INF));
	for (int i=0; i<=m; i++)
	{
		vector<vector<int>> diff(n+1, vector<int>(n+1, 0));
		for (int z1=0; z1<=n; z1++)
		{
			for (int z2=0; z2<=n; z2++)
			{
				for (int z=0; z<=m; z++) if (z!=i)
					diff[z1][z2]+=abs(a[z1][z]-a[z2][z]);
			}
		}
		vector<vector<int>> mn(1<<(n+1), vector<int>(n+1, INF));

		for (int j=0; j<=n; j++) mn[1<<j][j]=0;
		for (int msk=1; msk<(1<<(n+1)); msk++)
		{
			vector<int> setbits;
			for (int i=0; i<=n; i++) if ((1<<i)&msk) setbits.push_back(i);
			if ((int)setbits.size()<=1)
			{
				if (setbits.size()==n)
					for (int j2=0; j2<=n; j2++)
						if (!((1<<j2)&msk))
							ans[j2][i]=min(ans[j2][i], mn[msk][setbits[0]]);
				continue;
			}
			for (int j1=0; j1<=n; j1++) if ((1<<j1)&msk)
			{
				for (int j2=0; j2<=n; j2++) if ((1<<j2)&msk)
				{
					if (j1==j2) continue;
					mn[msk][j1]=min(mn[msk][j1], mn[msk^(1<<j1)][j2]+diff[j1][j2]);
				}
				if (setbits.size()==n)
					for (int j2=0; j2<=n; j2++)
						if (!((1<<j2)&msk))
							ans[j2][i]=min(ans[j2][i], mn[msk][j1]);
			}
		}
	}

	int res=INF;
	for (int i=0; i<=n; i++)
	{
		vector<vector<int>> diff(m+1, vector<int>(m+1, 0));
		for (int z1=0; z1<=m; z1++)
		{
			for (int z2=0; z2<=m; z2++)
			{
				for (int z=0; z<=n; z++) if (z!=i)
				diff[z1][z2]+=abs(a[z][z1]-a[z][z2]);
			}
		}

		vector<vector<int>> mn(1<<(m+1), vector<int>(m+1, INF));

		for (int j=0; j<=m; j++) mn[1<<j][j]=0;
		for (int msk=1; msk<(1<<(m+1)); msk++)
		{
			vector<int> setbits;
			for (int i=0; i<=m; i++) if ((1<<i)&msk) setbits.push_back(i);
			if ((int)setbits.size()<=1) continue;
			for (int j1=0; j1<=m; j1++) if ((1<<j1)&msk)
			{
				for (int j2=0; j2<=m; j2++) if ((1<<j2)&msk)
				{
					if (j1==j2) continue;
					mn[msk][j1]=min(mn[msk][j1], mn[msk^(1<<j1)][j2]+diff[j1][j2]);
				}
				if (setbits.size()==m)
				for (int j2=0; j2<=m; j2++)
				if (!((1<<j2)&msk))
				{
					res=min(res, ans[i][j2]+mn[msk][j1]);
				}
			}
		}
	}
	cout<<res<<'\n';
	return;
}

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(NULL); cout.tie(NULL);
	int t;
	cin >> t;
	//t=1;
	//cout<<'\n';
	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
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
```

This test input includes a variety of small-scale scenarios to test the algorithm's ability to handle different matrix sizes and values. The input is designed to cover edge cases such as a single row, a single column, and matrices with varying values that might expose issues in the XOR operations and the calculation of the beauty of the matrix.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
/*
考虑给该矩阵多补上一行一列。
原本同行同列 => 转化后同行同列
可以枚举丢掉哪一行哪一列。
因此可以行列分开考虑。这里考虑一下行。
n，m较小
假设丢掉第 i 行 j 列，我们需最小化其他 n 行两两差的绝对值的最小值。
可以使用状压 dp 解决。
复杂度好像是 $O(n \times 2^n)$
感觉上限不是很紧，我直接不精细实现
*/
using ll=long long;
void cmin(ll &x, ll y) {
	if(y<x) x=y;
}
const int N=18;
const ll inf=1e17;
ll f[1<<N][N];
int n, m, a[N][N];
ll valh(int x, int y, int d) {
	ll res=0;
	for(int i=1; i<=m; i++)
		if(d!=i)
			res+=abs(a[x][i]-a[y][i]);
	return res;
}
ll vall(int x, int y, int d) {
	ll res=0;
	for(int i=1; i<=n; i++)
		if(d!=i)
			res+=abs(a[i][x]-a[i][y]);
	return res;
}
void Sol() {
	scanf("%d %d", &n, &m);
	for(int i=1; i<=n; i++) {
		int x=0;
		for(int j=1; j<=m; j++) {
			scanf("%d", &a[i][j]);
			x^=a[i][j];
		}
		a[i][m+1]=x;
	} m++;
	for(int j=1; j<=m; j++) {
		int x=0;
		for(int i=1; i<=n; i++)  {
			x^=a[i][j];
		}
		a[n+1][j]=x;
	} n++;
	int S1=1<<n, S2=1<<m;
	ll ans=inf;
	for(int dh=1; dh<=n; dh++) {
		for(int dl=1; dl<=m; dl++) {

			ll res1=inf, res2=inf;
			memset(f, 0x3f, sizeof(f));
			for(int i=1; i<=n; i++)
				if(i!=dh) f[1<<(i-1)][i]=0;
			for(int s=1; s<S1; s++) {
				int lw=__lg(s&-s)+1;
				for(int j=1; j<=n; j++) {
					if(j==dh) continue ;
					if(f[s][j]<inf) {

						for(int k=lw+1; k<=n; k++) {
							if(k==dh) continue ;
							cmin(f[s^(1<<k-1)][k], f[s][j]+valh(j, k, dl));
						}
					}
				}
			}
			for(int i=1; i<=n; i++)
				if(i!=dh) cmin(res1, f[(S1-1)^(1<<dh-1)][i]);

			memset(f, 0x3f, sizeof(f));
			for(int i=1; i<=m; i++)
				if(i!=dl) f[1<<(i-1)][i]=0;
			for(int s=1; s<S2; s++) {
				int lw=__lg(s&-s)+1;
				for(int j=1; j<=m; j++) {
					if(j==dl) continue ;
					if(f[s][j]<inf) {

						for(int k=lw+1; k<=m; k++) {
							if(k==dl) continue ;

							cmin(f[s^(1<<k-1)][k], f[s][j]+vall(j, k, dh));
						}
					}
				}
			}
			for(int i=1; i<=m; i++)
				if(i!=dl)  cmin(res2, f[(S2-1)^(1<<dl-1)][i]);
//			if(res1+res2<ans) cout<<dh<<" "<<dl<<endl;
			cmin(ans, res1+res2);
		}
	}
	printf("%lld\n", ans);
}
int main() {
	int T; scanf("%d", &T);
	while(T--) Sol();
	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 expose the bug in the given code, we need to focus on edge cases or scenarios that can maximize the failure probability. The code uses dynamic programming to minimize the beauty of the matrix by considering the removal of one row and one column. Here are the key points to consider:

1. **Small Matrix with Repeated Values**: A matrix with repeated values can lead to incorrect XOR calculations.
2. **Matrix with Zero Values**: A matrix with zeros can cause issues in XOR calculations and absolute difference computations.
3. **Matrix with Large Values**: Large values can lead to overflow issues or incorrect XOR results.
4. **Matrix with Alternating Values**: Alternating values can lead to incorrect XOR results and absolute difference calculations.

Given these considerations, let's generate a small-scale test input that is complete and valid:

```plaintext
1
2 2
0 1
1 0
```

This test case includes a small 2x2 matrix with alternating values. This can help expose issues in the XOR calculations and the dynamic programming logic used to minimize the beauty of the matrix.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

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

using i64 = long long;

void solve() {
    int n, m;
    cin >> n >> m;
    vector<vector<int> > a(n + 1, vector<int>(m + 1));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> a[i][j];
            a[n][j] ^= a[i][j];

        }
    }
    vector<vector<int> > g0(n + 1, vector<int>(n + 1)), g1(m + 1, vector<int>(m + 1));

    auto work = [&](int n, vector<vector<int> > &g) -> int {
        int res = 2e9;
        vector<vector<int> > f(1 << n, vector<int>(n));
        for (int i = 1; i < (1 << n); i++) {
            for (int j = 0; j < n; j++) if (i >> j & 1) {
                int la = i ^ (1 << j);
                if (!la) break;
                f[i][j] = 2e9;
                for (int k = 0; k < n; k++) if (i >> k & 1) {
                    if (k == j) continue;
                    f[i][j] = min(f[i][j], g[j][k] + f[la][k]);
                }
            }
        }
        for (int i = 0; i < n; i++) res = min(res, f[(1 << n) - 1][i]);
        return res;
    };

    int ans = 2e9;
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= m; j++) {

            for (int k = 0; k < n; k++) {
                for (int l = 0; l < m; l++) {
                    if (i != n && k != i) a[i][l] ^= a[k][l];
                    if (j != m && j != l) a[k][j] ^= a[k][l];
                }
            }

            // for (int i = 0; i < n; i++) {
            //     for (int j = 0; j < m; j++) {
            //         cout << a[i][j] << " \n"[j == m - 1];
            //     }
            // }

            for (int k = 0; k < n; k++) {
                for (int l = 0; l < n; l++) {
                    g0[k][l] = 0;
                    for (int p = 0; p < m; p++) g0[k][l] += abs(a[k][p] - a[l][p]);
                }
            }
            for (int k = 0; k < m; k++) {
                for (int l = 0; l < m; l++) {
                    g1[k][l] = 0;
                    for (int p = 0; p < n; p++) g1[k][l] += abs(a[p][k] - a[p][l]);
                }
            }

            ans = min(ans, work(n, g0) + work(m, g1));
            // cout << i << ' ' << j << ' ' << work(n, g0) << ' ' << work(m, g1) << '\n';

            for (int k = 0; k < n; k++) {
                for (int l = 0; l < m; l++) {
                    if (i != n && k != i) a[i][l] ^= a[k][l];
                    if (j != m && j != l) a[k][j] ^= a[k][l];
                }
            }
        }
    }
    cout << ans << '\n';
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int T;
    cin >> T;
    while (T--) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

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

using i64 = long long;

void solve() {
    int n, m;
    cin >> n >> m;
    vector<vector<int> > a(n + 1, vector<int>(m + 1));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> a[i][j];
            a[n][j] ^= a[i][j];

        }
    }
    vector<vector<int> > g0(n + 1, vector<int>(n + 1)), g1(m + 1, vector<int>(m + 1));

    auto work = [&](int n, vector<vector<int> > &g) -> int {
        int res = 0x3f3f3f3f;
        vector<vector<int> > f(1 << n, vector<int>(n));
        for (int i = 1; i < (1 << n); i++) {
            for (int j = 0; j < n; j++) if (i >> j & 1) {
                int la = i ^ (1 << j);
                if (!la) break;
                f[i][j] = 0x3f3f3f3f;
                for (int k = 0; k < n; k++) if (i >> k & 1) {
                    if (k == j) continue;
                    f[i][j] = min(f[i][j], g[j][k] + f[la][k]);
                }
            }
        }
        for (int i = 0; i < n; i++) res = min(res, f[(1 << n) - 1][i]);
        return res;
    };

    int ans = 1e9;
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= m; j++) {

            for (int k = 0; k < n; k++) {
                for (int l = 0; l < m; l++) {
                    if (i != n && k != i) a[i][l] ^= a[k][l];
                    if (j != m && j != l) a[k][j] ^= a[k][l];
                }
            }

            // for (int i = 0; i < n; i++) {
            //     for (int j = 0; j < m; j++) {
            //         cout << a[i][j] << " \n"[j == m - 1];
            //     }
            // }

            for (int k = 0; k < n; k++) {
                for (int l = 0; l < n; l++) {
                    g0[k][l] = 0;
                    for (int p = 0; p < m; p++) g0[k][l] += abs(a[k][p] - a[l][p]);
                }
            }
            for (int k = 0; k < m; k++) {
                for (int l = 0; l < m; l++) {
                    g1[k][l] = 0;
                    for (int p = 0; p < n; p++) g1[k][l] += abs(a[p][k] - a[p][l]);
                }
            }

            ans = min(ans, work(n, g0) + work(m, g1));
            // cout << i << ' ' << j << ' ' << work(n, g0) << ' ' << work(m, g1) << '\n';

            for (int k = 0; k < n; k++) {
                for (int l = 0; l < m; l++) {
                    if (i != n && k != i) a[i][l] ^= a[k][l];
                    if (j != m && j != l) a[k][j] ^= a[k][l];
                }
            }
        }
    }
    cout << ans << '\n';
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int T;
    cin >> T;
    while (T--) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

Buggy Code:
from functools import reduce, cache
import operator
import sys
input = sys.stdin.readline
MOD = 10**9 + 7

def read_array(factory):
    return [factory(num) for num in input().strip().split()]
def print_array(arr):
    print(" ".join(map(str, arr)))

def solve(mat):
    for i in mat:
        i.append(reduce(operator.xor, i))
    n = []
    for i in range(len(mat[0])):
        n.append(reduce(operator.xor, [mat[j][i] for j in range(len(mat))]))
    mat.append(n)
    # Swap any rows of mat greedily
    def diff(r1, r2):
        return sum(abs(i-j) for i, j in zip(r1, r2))

    def had(mask, last):
        dp = {(0, None): 0}
        diffs = {(l, i): diff(mat2[l], mat2[i]) for l in range(len(mat2)) for i in range(len(mat2))}
        for _ in range(len(mat2)-1):
            newdp = {}
            for (mask, last), val in dp.items():
                for i in range(len(mat2)):
                    if mask & (1 << i) == 0:
                        newmask = mask ^ (1 << i)
                        if last == None:
                            newdp[(newmask, i)] = min(val, newdp.get((newmask, i), float('inf')))
                        else:
                            # m = min(m, had(mask ^ (1 << i), i) + diff(mat2[last], mat2[i]))
                            newdp[(newmask, i)] = min(val + diffs[(last, i)], newdp.get((newmask, i), float('inf')))
            dp = newdp
        ans = {}
        for (mask, last), val in dp.items():
            for i in range(len(mat2)):
                if mask & 1 << i == 0:
                    ans[i] = min(ans.get(mask, float('inf')), val)
        return ans

    m = float('inf')
    ans2 = []
    for omit_col in range(len(mat[0])):
        mat2 = [[mat[i][j] for j in range(len(mat[0])) if j != omit_col] for i in range(len(mat))]
        ans2.append(had(0, None))

    ans3 = []
    for omit_row in range(len(mat)):
        mat2 = [[mat[i][j] for j in range(len(mat[0]))] for i in range(len(mat)) if i != omit_row]
        mat2 = [[mat2[j][i] for j in range(len(mat2))] for i in range(len(mat2[0]))]
        ans3.append(had(0, None))

    m = float('inf')
    for omit_row in range(len(mat)):
        for omit_col in range(len(mat[0])):
            m = min(m, ans2[omit_col][omit_row] + ans3[omit_row][omit_col])
            # print(ans2)
            # print(ans3)
    return m

if __name__ == "__main__":
    # Read in input:
    t = int(input())

    # Call solution:
    for _ in range(t):
        n, _ = read_array(int)
        mat = []
        for _ in range(n):
            mat.append(read_array(int))
        ans = solve(mat)
        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 1
5
2 2
1 2
3 4
3 3
1 0 1
0 1 0
1 0 1
4 4
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

Buggy Code:
T=int(input())
for t in range(T):
    N,M=map(int,input().split())
    A=[list(map(int,input().split())) for n in range(N)]
    for n in range(N):
        A[n].append(0)
        for m in range(M):
            A[n][-1]^=A[n][m]
    A.append([0]*(M+1))
    for n in range(N):
        for m in range(M+1):
            A[N][m]^=A[n][m]
    def solve(x,y):
        B=[[A[i][j] for j in range(M+1) if j!=y] for i in range(N+1) if i!=x]
        C=[[sum(abs(B[n][m0]-B[n][m1]) for n in range(N)) for m1 in range(M)] for m0 in range(M)]
        D=[[sum(abs(B[n0][m]-B[n1][m]) for m in range(M)) for n1 in range(N)] for n0 in range(N)]
        def solve1(E):
            M=len(E)
            inf=1<<60
            dp=[[inf]*M for bit in range(1<<M)]
            ans=inf
            for bit in range(1,1<<M):
                idx=[m for m in range(M) if bit&1<<m]
                if len(idx)==1:
                    m=idx[0]
                    dp[bit][m]=0
                else:
                    for m1 in idx:
                        for m0 in idx:
                            if m0==m1:
                                continue
                            dp[bit][m1]=min(dp[bit][m1],dp[bit^1<<m1][m0]+E[m0][m1])
                if len(idx)==M:
                    ans=min(ans,min(dp[bit]))
            return ans
        ans=solve1(C)+solve1(D)
        return ans
    ans=min(solve(x,y) for x in range(N+1) for y in range(N+1))
    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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The buggy code seems to have issues with how it handles the XOR operations and the dynamic programming part. Let's focus on a small-scale test input that can highlight these issues.

### Step-by-Step Reasoning:

1. **Matrix Size**: Start with a small matrix size, such as 3x3, to ensure the input is manageable and easy to analyze.
2. **Matrix Content**: Use values that will produce interesting XOR results, such as powers of 2, to ensure that the XOR operations have a significant impact.
3. **Edge Cases**: Consider matrices with all elements being the same, or matrices with alternating values to test the XOR operations and the dynamic programming logic.

### Chosen Test Input:
Let's use a 3x3 matrix with alternating values to test the XOR operations and the dynamic programming logic.

```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
```

### Explanation:
- **Matrix Size**: 3x3, which is small enough to be manageable.
- **Matrix Content**: Alternating values from 1 to 9, which will produce interesting XOR results.
- **Edge Cases**: This matrix will test the XOR operations on rows and columns and the dynamic programming part to ensure it correctly calculates the minimum beauty.

This input should help expose any issues in the buggy code, particularly in how it handles the XOR operations and the dynamic programming logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

Buggy Code:
from functools import reduce, cache
import operator
import sys
input = sys.stdin.readline
MOD = 10**9 + 7

def read_array(factory):
    return [factory(num) for num in input().strip().split()]
def print_array(arr):
    print(" ".join(map(str, arr)))

def solve(mat):
    for i in mat:
        i.append(reduce(operator.xor, i))
    n = []
    for i in range(len(mat[0])):
        n.append(reduce(operator.xor, [mat[j][i] for j in range(len(mat))]))
    mat.append(n)
    # Swap any rows of mat greedily
    def diff(r1, r2):
        return sum(abs(i-j) for i, j in zip(r1, r2))

    def had(mask, last):
        dp = {(0, None): 0}
        diffs = {(l, i): diff(mat2[l], mat2[i]) for l in range(len(mat2)) for i in range(len(mat2))}
        for _ in range(len(mat2)):
            newdp = {}
            for (mask, last), val in dp.items():
                for i in range(len(mat2)):
                    if mask & (1 << i) == 0:
                        newmask = mask ^ (1 << i)
                        if last == None:
                            newdp[(newmask, i)] = min(val, newdp.get((newmask, i), float('inf')))
                        else:
                            # m = min(m, had(mask ^ (1 << i), i) + diff(mat2[last], mat2[i]))
                            newdp[(newmask, i)] = min(val + diffs[(last, i)], newdp.get((newmask, i), float('inf')))
            dp = newdp
        return min(dp.values())

    m = float('inf')
    for omit_row in range(len(mat)):
        for omit_col in range(len(mat[0])):
            # print(omit_row, omit_col)
            mat2 = [[mat[i][j] for j in range(len(mat[0])) if j != omit_col] for i in range(len(mat)) if i != omit_row]
            x1 = had(0, None)
            mat2 = [[mat2[j][i] for j in range(len(mat2))] for i in range(len(mat2[0]))]
            x2 = had(0, None)
            m = min(m, x1 + x2)
    return m

if __name__ == "__main__":
    # Read in input:
    t = int(input())

    # Call solution:
    for _ in range(t):
        n, _ = read_array(int)
        mat = []
        for _ in range(n):
            mat.append(read_array(int))
        ans = solve(mat)
        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
1
2 2
0 1
1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

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

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

int n,m;

int bannedRow,bannedCol;

ll memoRow[1<<16][16];
ll memoCol[1<<16][16];

ll diffRow[16][16];
ll diffCol[16][16];

vector<vi> grid(16,vi(16,0));

ll dpRow(int curr, int mask){
	if ((mask|(1<<bannedRow)) == (1<<(n+1))-1) return 0;
	if (memoRow[mask][curr] != -1) return memoRow[mask][curr];

	ll best = 1e9;
	rep(nxt,0,n+1){
		if (nxt == bannedRow) continue;
		if (((mask>>nxt)&1) == 0){
			best = min(best,dpRow(nxt,mask|1<<nxt) + diffRow[nxt][curr] - abs(grid[curr][bannedCol] - grid[nxt][bannedCol]));
		}
	}

	memoRow[mask][curr] = best;

	return best;
}

ll dpCol(int curr, int mask){
	//cout << curr << " " << mask << "\n";
	if ((mask|(1<<bannedCol)) == ((1<<(m+1))-1)) return 0;
	if (memoCol[mask][curr] != -1) return memoCol[mask][curr];

	//cout << curr << " " << mask << "\n";

	ll best = 1e9;
	rep(nxt,0,m+1){
		if (nxt == bannedCol) continue;
		if (((mask>>nxt)&1) == 0){
			best = min(best,dpCol(nxt,mask|1<<nxt) + diffCol[nxt][curr] - abs(grid[bannedRow][nxt] - grid[bannedRow][curr]));
		}
	}

	memoCol[mask][curr] = best;

	//cout << best << " "<< curr << " " << mask << "\n";

	return best;
}

int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin.exceptions(cin.failbit);

	int t;
	cin >> t;
	rep(testcase,0,t){
		cin >> n >> m;

		rep(i,0,n) rep(j,0,m) cin >> grid[i][j];

		rep(i,0,n){
			int temp = 0;
			rep(j,0,m) temp ^= grid[i][j];
			grid[i][m] = temp;
		}

		rep(j,0,m+1){
			int temp = 0;
			rep(i,0,n){
				temp ^= grid[i][j];
			}
			grid[n][j] = temp;
		}

		rep(from,0,n+1){
			rep(to,from+1,n+1){
				ll temp = 0;
				rep(i,0,m+1){
					temp += llabs(grid[from][i] - grid[to][i]);
				}

				diffRow[from][to] = temp;
				diffRow[to][from] = temp;
			}
		}

		rep(from,0,m+1){
			rep(to,from+1,m+1){
				ll temp = 0;
				rep(i,0,n+1){
					temp += llabs(grid[i][from] - grid[i][to]);
				}
				diffCol[from][to] = temp;
				diffCol[to][from] = temp;
			}
		}

		ll best = 1e9;

		rep(i,0,n+1) rep(j,0,m+1){
			bannedRow = i;
			bannedCol = j;

			rep(x,0,1<<16) rep(y,0,16) {
				memoRow[x][y] = -1;
				memoCol[x][y] = -1;
			}

			ll currRow = 1e9;
			ll currCol = 1e9;
			rep(row,0,n+1){
				if (row == bannedRow) continue;
				currRow = min(currRow,dpRow(row,1<<row));
			}

			rep(col,0,m+1){
				if (col == bannedCol) continue;
				currCol = min(currCol,dpCol(col,1<<col));
			}
			//cout << "\n";
			best = min(currCol + currRow,best);
		}

		cout << best << "\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:
To create a test input that exposes the bug in the provided code, we need to focus on scenarios where the dynamic programming (DP) approach might fail. Specifically, we should look at cases where the DP states and transitions might not be correctly handled.

### Key Considerations:
1. **Small Matrix Size**: Since the matrix size is limited to \(15 \times 15\), we should use a small matrix to ensure the input is valid and fits within the constraints.
2. **Edge Cases**: Consider matrices with very low or very high values, or matrices that are almost uniform but have a few different values.
3. **Symmetry and Asymmetry**: Test both symmetric and asymmetric matrices to see how the DP handles different structures.
4. **Corner Cases**: Test matrices where the initial values are already optimal or nearly optimal.

### Test Input:
Let's create a \(3 \times 3\) matrix with values that might cause issues in the DP transitions. We will use a matrix with a mix of low and high values to see if the DP correctly handles the transitions.

```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
```

### Explanation:
- **Matrix Size**: \(3 \times 3\) to ensure it is small and manageable.
- **Values**: The matrix contains values from 1 to 9, which are not uniform and have a clear pattern. This can help expose issues in the DP transitions.
- **Expected Output**: The expected output for this input is 24, which is the minimum beauty for the given matrix after applying the operations.

This test input should help in identifying any issues in the DP logic, especially in how it handles the transitions and the memoization.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

Buggy Code:
// #pragma optimize("Ofast")
#include "bits/stdc++.h"
#define all(v) (v).begin(), (v).end()
#define pb push_back
#define em emplace_back
#define mp make_pair
#define F first
#define S second

using namespace std;
template<class C>
using vec = vector<C>;
using vi = vector<int>;
using vpi = vector<pair<int, int>>;
using pii = pair<int, int>;

int calc(int n, int m, vec<vi> a) {
  int dp[1 << m][m];
  for (int i = 0; i < (1 << m); i++)
    for (int j = 0; j < m; j++)
      dp[i][j] = INT_MAX;
  for (int i = 0; i < m; i++) {
    dp[1 << i][i] = 0;
  }
  int res = INT_MAX;
  for (int mask = 1; mask < (1 << m); mask++) {
    if (__builtin_popcount(mask) == m - 1) {
      for (int b = 0; b + 1 < m; b++) {
        if ((mask >> b) & 1) {
          res = min(res, dp[mask][b]);
        }
      }
    }
    for (int i = 0; i + 1 < m; i++) {
      if (!((mask >> i) & 1)) continue;
      for (int to = 0; to < m; to++) {
        if ((mask >> to) & 1) continue;
        int relax = dp[mask][i];
        if (relax == INT_MAX) continue;
        for (int j = 1; j < n; j++) {
          relax += abs(a[j][i] - a[j][to]);
        }
        dp[mask ^ (1 << to)][to] = min(dp[mask ^ (1 << to)][to], relax);
      }
    }
  }
  return res;
}

void solve() {
  int n, m;
  cin >> n >> m;
  vec<vi> a(n + 1, vi(m + 1));
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
      a[i].back() ^= a[i][j];
      a[0][j] ^= a[i][j];
      a[0].back() ^= a[i][j];
    }
  }
  int res = INT_MAX;
  for (int z = 0; z <= n; z++) {
    for (int z2 = 0; z2 <= m; z2++) {
      swap(a[z], a[0]);
      for (int i = 0; i <= n; i++) {
        swap(a[i][z2], a[i].back());
      }
      int ans = 0;
      ans += calc(n + 1, m + 1, a);
      vec<vi> b(m + 1, vi(n + 1));
      for (int j = m; j >= 0; j--) {
        for (int i = n; i >= 0; i--) {
          b[m - j][n - i] = a[i][j];
        }
      }
      ans += calc(m + 1, n + 1, b);
      res = min(res, ans);
      for (int i = 0; i <= n; i++) {
        swap(a[i][z2], a[i].back());
      }
      swap(a[z], a[0]);
    }
  }
  cout << res << '\n';
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Matrix Size**: Since the matrix size is limited to \(1 \le n, m \le 15\), we can start with smaller matrices and gradually increase the size to see if the bug becomes more apparent.

2. **Matrix Values**: The values in the matrix are non-negative integers less than \(2^{20}\). We should consider using values that can lead to interesting XOR results, such as powers of 2, zero, and repeated values.

3. **Symmetry and Patterns**: Matrices with symmetry or specific patterns can sometimes expose bugs in algorithms that do not handle these cases correctly.

4. **Corner Cases**: Consider matrices with all zeros, matrices with a single non-zero value, and matrices where all values are the same.

5. **Complexity**: Matrices with a mix of values that require multiple operations to achieve the minimum beauty can also be challenging.

Let's create a test input that includes a matrix with a mix of values and a specific pattern to see if the bug becomes apparent.

### Chosen Test Input
We will use a \(3 \times 3\) matrix with a mix of values and a pattern that might expose the bug.

```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
```

### Explanation
- **Matrix Size**: \(3 \times 3\) to keep it small and manageable.
- **Matrix Values**: A simple increasing sequence from 1 to 9.
- **Pattern**: The values form a simple arithmetic progression, which can be challenging for algorithms that do not handle linear patterns well.

This test input is designed to be simple enough to understand but complex enough to potentially expose any issues in the given code. The expected output for this test case is the minimum possible beauty of the matrix after applying the operations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

Buggy Code:
T=int(input())
for t in range(T):
    N,M=map(int,input().split())
    A=[list(map(int,input().split())) for n in range(N)]
    for n in range(N):
        A[n].append(0)
        for m in range(M):
            A[n][-1]^=A[n][m]
    A.append([0]*(M+1))
    for n in range(N):
        for m in range(M+1):
            A[N][m]^=A[n][m]
    def solve(x,y):
        B=[[A[i][j] for j in range(M+1) if j!=y] for i in range(N+1) if i!=x]
        C=[[sum(abs(B[n][m0]-B[n][m1]) for n in range(N)) for m1 in range(M)] for m0 in range(M)]
        D=[[sum(abs(B[n0][m]-B[n1][m]) for m in range(M)) for n1 in range(N)] for n0 in range(N)]
        def solve1(E):
            M=len(E)
            inf=1<<60
            dp=[[inf]*M for bit in range(1<<M)]
            ans=inf
            for bit in range(1,1<<M):
                idx=[m for m in range(M) if bit&1<<m]
                if len(idx)==1:
                    m=idx[0]
                    dp[bit][m]=0
                else:
                    for m1 in idx:
                        for m0 in idx:
                            if m0==m1:
                                continue
                            dp[bit][m1]=min(dp[bit][m1],dp[bit^1<<m1][m0]+E[m0][m1])
                if len(idx)==M:
                    ans=min(ans,min(dp[bit]))
            return ans
        ans=solve1(C)+solve1(D)
        return ans
    ans=min(solve(x,y) for x in range(N+1) for y in range(M+1))
    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
1
3 3
1 2 3
4 5 6
7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

Buggy Code:
ans = []
inf = 10**12
def salesman(dists):
	N = len(dists)
	v = [[inf]*(1<<N) for _ in range(N)]

	for sta in range(N):
		v[sta][1<<sta] = 0

	for mask in range(1<<N):
		for sta in range(N):
			if not (mask>>sta)&1: continue
			for end in range(N):
				if (mask>>end)&1: continue
				v[end][mask|(1<<end)] = min( v[end][mask|(1<<end)], v[sta][mask]+dists[sta][end] )

	#print(v)
	return min( v[sta][(1<<N)-1] for sta in range(N) )

for _ in range(int(input())):
	n,m = [int(x) for x in input().split()]
	g = [[int(x) for x in input().split()] for _ in range(n)]

	h = [[0]*(m+1) for _ in range(n+1)]
	for y in range(n):
		for x in range(m):
			h[y][x] = g[y][x]
			h[n][x] ^= g[y][x]
			h[y][m] ^= g[y][x]
			h[n][m] ^= g[y][x]

	bb = inf
	for covery in range(n+1):
		for coverx in range(m+1):
			dists = []
			for y1 in range(n+1):
				if y1 == covery: continue
				line = []
				for y2 in range(n+1):
					if y2 == covery: continue
					line.append( sum( abs(h[y2][x]-h[y1][x]) for x in range(m+1) if not x==coverx ) )
				dists.append(line)
			b = salesman(dists)

			dists = []
			for x1 in range(m+1):
				if x1 == coverx: continue
				line = []
				for x2 in range(m+1):
					if x2 == coverx: continue
					line.append( sum( abs(h[y][x2]-h[y][x1]) for y in range(n+1) if not y==covery ) )
				dists.append(line)
			b += salesman(dists)
			bb = min(b,bb)
	ans.append(bb)

for w in ans:
	print(w)

**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 for the Xor-Grid Problem, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to think through the process:

1. **Understand the Buggy Code**:
   - The buggy code attempts to solve the problem using a dynamic programming approach similar to the Traveling Salesman Problem (TSP), but it is not correctly handling the XOR operations and the beauty calculation.
   - The code defines a `salesman` function to find the minimum path in a distance matrix, which is not directly applicable to the Xor-Grid Problem.

2. **Identify Potential Issues**:
   - The `salesman` function is not designed to handle the XOR operations and the beauty calculation.
   - The code does not correctly compute the beauty of the matrix after applying the operations.
   - The code may fail to handle matrices with specific patterns or edge cases.

3. **Generate a Test Input**:
   - We need a small-scale test input that can expose these issues.
   - Consider a matrix that has a specific pattern or structure that can cause the buggy code to fail.

4. **Test Input**:
   - A 2x2 matrix with all elements being 0 can be a simple edge case.
   - A 2x2 matrix with all elements being the same non-zero value can also be a simple edge case.
   - A 3x3 matrix with a specific pattern that requires multiple operations to achieve the minimum beauty.

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

```plaintext
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
```

However, to focus on a more specific edge case, let's use a 2x2 matrix with all elements being 0:

```plaintext
1
2 2
0 0
0 0
```

This test input will help expose the bug because the buggy code might not correctly handle a matrix with all elements being 0, which should result in a beauty of 0. If the buggy code fails to return 0, it indicates a significant issue in handling such simple cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Xor-Grid Problem
time_limit_ms: 5000
memory_limit_mb: 256
Description: Given a matrix $a$ of size $n \times m$, each cell of which contains a non-negative integer. The integer lying at the intersection of the $i$-th row and the $j$-th column of the matrix is called $a_{i,j}$.

Let's define $f(i)$ and $g(j)$ as the [XOR](https://en.wikipedia.org/wiki/Exclusive_or) of all integers in the $i$-th row and the $j$-th column, respectively. In one operation, you can either:

  * Select any row $i$, then assign $a_{i,j} := g(j)$ for each $1 \le j \le m$; or    * Select any column $j$, then assign $a_{i,j} := f(i)$ for each $1 \le i \le n$.

![](CDN_BASE_URL/3bba110245afb461968dbb618fc60828) An example of applying an operation on column $2$ of the matrix.

In this example, as we apply an operation on column $2$, all elements in this column are changed:

  * $a_{1,2} := f(1) = a_{1,1} \oplus a_{1,2} \oplus a_{1,3} \oplus a_{1,4} = 1 \oplus 1 \oplus 1 \oplus 1 = 0$    * $a_{2,2} := f(2) = a_{2,1} \oplus a_{2,2} \oplus a_{2,3} \oplus a_{2,4} = 2 \oplus 3 \oplus 5 \oplus 7 = 3$    * $a_{3,2} := f(3) = a_{3,1} \oplus a_{3,2} \oplus a_{3,3} \oplus a_{3,4} = 2 \oplus 0 \oplus 3 \oplus 0 = 1$    * $a_{4,2} := f(4) = a_{4,1} \oplus a_{4,2} \oplus a_{4,3} \oplus a_{4,4} = 10 \oplus 11 \oplus 12 \oplus 16 = 29$

You can apply the operations any number of times. Then, we calculate the $\textit{beauty}$ of the final matrix by summing the absolute differences between all pairs of its adjacent cells.

More formally, $\textit{beauty}(a) = \sum|a_{x,y} - a_{r,c}|$ for all cells $(x, y)$ and $(r, c)$ if they are adjacent. Two cells are considered adjacent if they share a side.

Find the minimum $\textit{beauty}$ among all obtainable matrices.

Input: The first line contains a single integer $t$ ($1 \le t \le 250$) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ($1 \le n, m \le 15$) — the number of rows and columns of $a$, respectively.

The next $n$ lines, each containing $m$ integers $a_{i,1}, a_{i,2}, \ldots, a_{i,m}$ ($0 \le a_{i,j} < 2^{20}$) — description of the matrix $a$.

It is guaranteed that the sum of $(n^2 + m^2)$ over all test cases does not exceed $500$.

Output: For each test case, print a single integer $b$ — the smallest possible $\textit{beauty}$ of the matrix.

Examples:
input:
4
1 2
1 3
2 3
0 1 0
5 4 4
2 3
0 2 4
4 5 1
3 3
1 2 3
4 5 6
7 8 9
output:
1
3
13
24

Note: Let's denote $r(i)$ as the first type operation applied on the $i$-th row, and $c(j)$ as the second type operation applied on the $j$-th column.

In the first test case, you can apply an operation $c(1)$, which assigns $a_{1,1} := 1 \oplus 3 = 2$. Then, we'll receive this matrix:

2| 3   ---|---      In the second test case, you can apply an operation $r(1)$, which assigns:

  * $a_{1,1} := g(1) = 0 \oplus 5 = 5$    * $a_{1,2} := g(2) = 1 \oplus 4 = 5$    * $a_{1,3} := g(3) = 0 \oplus 4 = 4$

The resulting matrix after performing the operation is:

5| 5| 4   ---|---|---   5| 4| 4      In the third test case, the best way to achieve minimum $\textit{beauty}$ is applying three operations: $c(3)$, $r(2)$, and $c(2)$. The resulting matrix is:

0| 4| 6   ---|---|---   4| 5| 6

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

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x...) 228
#endif

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

#define fi first
#define se second

const int inf = int(1e9) + 1;
const ll infll = ll(1e18) + 1;
int mod = int(1e9) + 7;
mt19937_64 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());

template<typename T1, typename T2>
bool chmin(T1 &a, T2 b) {
    if (a > b) return a = b, true;
    return false;
}
template<typename T1, typename T2>
bool chmax(T1 &a, T2 b) {
    if (a < b) return a = b, true;
    return false;
}

void solve_case() {
    int n, m; cin >> n >> m;
    vector<vector<int>> a(n + 1, vector<int>(m + 1));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> a[i][j];
            a[i][m] ^= a[i][j];
            a[n][j] ^= a[i][j];
            a[n][m] ^= a[i][j];
        }
    }
    ll ans = infll;
//    vector<vector<int>> all_pn, all_pm;
//    {
//        vector<int> perm(n + 1); iota(perm.begin(), perm.end(), 0);
//        do {
//            all_pn.push_back(perm);
//        } while (next_permutation(perm.begin(), perm.end()));
//    }
//    {
//        vector<int> perm(m + 1); iota(perm.begin(), perm.end(), 0);
//        do {
//            all_pm.push_back(perm);
//        } while (next_permutation(perm.begin(), perm.end()));
//    }
//    for (auto pn : all_pn) {
//        for (auto pm : all_pm) {
//            ll res = 0;
//            for (int i = 0; i + 1 < n; i++) {
//                for (int j = 0; j < m; j++) {
//                    res += abs(a[pn[i]][pm[j]] - a[pn[i + 1]][pm[j]]);
//                }
//            }
//            for (int i = 0; i < n; i++) {
//                for (int j = 0; j + 1 < m; j++) {
//                    res += abs(a[pn[i]][pm[j]] - a[pn[i]][pm[j + 1]]);
//                }
//            }
//            ans = min(ans, res);
//        }
//    }
//    cout << ans << "\n";
//    return;
    for (int vi = 0; vi <= n; vi++) {
        for (int vj = 0; vj <= m; vj++) {
            vector<vector<int>> na(n, vector<int>(m));
            int ptr_i = 0;
            for (int i = 0; i <= n; i++) {
                if (i == vi) continue;
                int ptr_j = 0;
                for (int j = 0; j <= m; j++) {
                    if (j == vj) continue;
                    na[ptr_i][ptr_j] = a[i][j];
                    ptr_j++;
                }
                ptr_i++;
            }
            swap(a, na);
            auto solve = [&]() -> ll {
                vector<vector<ll>> dp(1 << n, vector<ll>(n, infll));
                for (int mask = 1; mask < (1 << n); mask++) {
                    if (__builtin_popcount(mask) == 1) {
                        for (int i = 0; i < n; i++) {
                            if ((mask >> i) & 1) {
                                dp[mask][i] = 0;
                            }
                        }
                        continue;
                    }
                    for (int i = 0; i < n; i++) {
                        if (!((mask >> i) & 1)) continue;
                        for (int pi = 0; pi < n; pi++) {
                            if (pi == i || !((mask >> pi) & 1)) continue;
                            ll sum = 0;
                            for (int j = 0; j < m; j++) {
                                sum += abs(a[i][j] - a[pi][j]);
                            }
                            chmin(dp[mask][i], dp[mask ^ (1 << i)][pi] + sum);
                        }
                    }
                }
                return *min_element(dp.back().begin(), dp.back().end());
            };
            ll cur = solve();
            vector<vector<int>> b(m, vector<int>(n));
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    b[j][i] = a[i][j];
                }
            }
            swap(n, m);
            swap(a, b);
            cur += solve();
            swap(a, b);
            swap(n, m);
            swap(a, na);
            ans = min(ans, cur);
        }
    }
    cout << ans << "\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
#ifdef LOCAL
    clock_t tStart = clock();
#endif
    int T = 1; cin >> T;
    for (int test = 1; test <= T; test++) {
        solve_case();
    }
#ifdef LOCAL
    fprintf(stderr, "\n>> Runtime: %.10fs\n", (double) (clock() - tStart) / CLOCKS_PER_SEC);
#endif
}

**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 are likely to cause issues. The buggy code involves complex operations and dynamic programming, so we should focus on inputs that can lead to incorrect state transitions or incorrect calculations.

### Key Points to Consider:
1. **Matrix Size**: The matrix size should be small enough to be manageable but large enough to potentially expose bugs. A 3x3 or 4x4 matrix is a good starting point.
2. **Initial Values**: The initial values in the matrix should be chosen to create non-trivial XOR results, which can lead to incorrect state transitions.
3. **Edge Cases**: Consider matrices with all zeros, all ones, or alternating values to test edge cases.

### Chosen Test Input:
Let's choose a 3x3 matrix with values that create non-trivial XOR results and are not trivially zero.

```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
```

### Explanation:
- **Matrix Size**: 3x3, which is small but non-trivial.
- **Initial Values**:
  - The first row: 1, 2, 3
  - The second row: 4, 5, 6
  - The third row: 7, 8, 9

These values ensure that the XOR results for rows and columns are not trivially zero, which can help in testing the correctness of the XOR operations and the dynamic programming logic.

### Expected Behavior:
- The algorithm should correctly calculate the XOR values for each row and column.
- The dynamic programming should correctly handle the state transitions and minimize the beauty of the matrix.

### Why This Input is Challenging:
- **Non-trivial XOR Results**: The XOR results for rows and columns are not trivially zero, which can help in testing the correctness of the XOR operations.
- **Dynamic Programming**: The dynamic programming logic should correctly handle the state transitions, which can be complex with non-trivial values.
- **Edge Cases**: The input is small enough to be manageable but large enough to potentially expose bugs in the algorithm.

This input should help in identifying any issues in the buggy code.