**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

Buggy Code:
import java.util.Scanner;

public class corners {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t=scan.nextInt();
        while(t-->0)
        {
            int n=scan.nextInt();
            int m=scan.nextInt();
            String a[]=new String[n];
            String b[]=new String[n];
            for(int i=0;i<n;i++)
            {
                a[i]=scan.next();
            }
            for(int i=0;i<n;i++)
            {
                b[i]=scan.next();
            }
            int flag=0;
            for(int i=0;i<m;i++)
            {
                int suma=0;
                int sumb=0;
                for(int j=0;j<n;j++)
                {
                    suma+=(int)(a[j].charAt(i)-48);
                    sumb+=(int)(b[j].charAt(i)-48);
                }
                if(suma%3!=sumb%3)
                {
                    flag=1;
                    System.out.println("NO");
                    break;
                }
            }
            if(flag==0)
            {
                System.out.println("YES");
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
2
2 2
00
00
11
11
2 2
00
00
22
22
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

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

#define fast                 \
	ios::sync_with_stdio(0); \
	cin.tie(0);              \
	cout.tie(0);
#define pp pair<int,int>
#define ll long long int
#define ppll pair<ll,ll>
#define fs first
#define ss second

const ll mod=1e9+7;
ll maxll(ll a,ll b){return a>b?a:b;}
ll minll(ll a,ll b){return a<b?a:b;}
int gcd(int a, int b){
    return (a==0)?b:gcd(b % a, a);
}
ll gcd(ll a,ll b){
	return a==0?b:gcd(b%a,a);
}
ll powp(ll a, ll b){
	ll ans=1;
	while(b>0){
		if(b%2){
			ans=(ans*a)%mod;
			b--;
		}
		a=(a*a)%mod;
		b/=2;

	}
	return ans;
}
int invertBits(int n) {
    return (~n) & ((1LL << 31) - 1);
}

bool perfectsq(ll x){
	ll t=sqrt(x);
	if(t*t==x)return true;
	return false;
}

signed main()
{
	//#ifndef ONLINE_JUDGE
	//  freopen(“input.txt”, “r”, stdin);
	// freopen(“output.txt”, “w”, stdout);
	//#endif
	//(A+B-1)/B->ceil
	fast;
	// solution code
	int _t=1;
	cin>>_t;

	while(_t--){
		ll n,q,x=0,y=0,m,cnt=0,sum=0,f=1,idx=0,x1=0,x2=0,y1,y2,k=0;
		cin>>n>>m;
		vector<vector<int>> a(n,vector<int>(m)),b(n,vector<int>(m));

		for(auto&v:a){
			string x;
			cin>>x;
			for(int i=0;i<m;i++){
				v[i]=x[i]-'0';
			}
		}
		for(auto&v:b){
			string x;
			cin>>x;
			for(int i=0;i<m;i++){
				v[i]=x[i]-'0';
			}
		}
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
				a[i][j]=(b[i][j]-a[i][j]+3)%3;
		}}
		vector<int> v1,v2;
		for(int i=0;i<n;i++){
			int cnt1=0,cnt2=0;
			for(int j=0;j<m;j++)
				if(a[i][j]==1)cnt1++;
				else if(a[i][j]==2)cnt2++;

			k=(cnt1-cnt2);
			if((k!=0 && k%2==0)||k==1){f=0;break;}
			if(k!=0)v1.push_back(k);
			sum+=k;
		}

		for(int j=0;j<m;j++){
			int cnt1=0,cnt2=0;
			for(int i=0;i<n;i++)
				if(a[i][j]==1)cnt1++;
				else if(a[i][j]==2)cnt2++;

			k=(cnt1-cnt2);
			if((k!=0 && k%2==0)||k==1){f=0;break;}
			if(k!=0)v2.push_back(k);
			sum-=k;
		}

		if(sum!=0)f=0;
		if(!f){
			cout<<"NO\n";
			continue;
		}
		cout<<"YES\n";
	}
}
// n*n-n-2c
//(1+sqrt(-1+4*2*c))/2
//1111
//1101---
//1011

//  |
//1 3 7
//

**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
000
000
000
111
111
111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

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

bool canTransform(vector<vector<int>>& a, vector<vector<int>>& b, int n, int m) {
    for (int i = 0; i < n-1; i++) {
        for (int j = 0; j < m-1; j++) {
            if (a[i][j] != b[i][j]) {
                int diff = (b[i][j] - a[i][j] + 3) % 3;
                a[i][j] = b[i][j];
                a[i+1][j] = (a[i+1][j] + 2 * diff) % 3;
                a[i][j+1] = (a[i][j+1] + 2 * diff) % 3;
                a[i+1][j+1] = (a[i+1][j+1] + diff) % 3;
            }
        }
    }
    for (int i = 0; i < n; i++) {
        if (a[i][0] != b[i][0]) return false;
    }
    for (int j = 0; j < m; j++) {
        if (a[0][j] != b[0][j]) return false;
    }
    return true;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;
        vector<vector<int>> a(n, vector<int>(m));
        vector<vector<int>> b(n, vector<int>(m));

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                char c;
                cin >> c;
                a[i][j] = c - '0';
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                char c;
                cin >> c;
                b[i][j] = c - '0';
            }
        }

        if (canTransform(a, b, n, m)) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3 3
000
000
000
011
101
110
4 4
0000
0000
0000
0000
0012
0021
0100
0200
2 2
00
00
11
11
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

Buggy Code:
from sys import stdin
from collections import defaultdict,Counter
input = lambda:stdin.readline()[:-1]
import math
for _ in range(int(input())):
	# n = int(input())
	n,m = map(int,input().split())
	# a = list(map(int,input().split()))
	a,b = [],[]
	for i in range(n):
		a.append(list(map(int,list(input()))))
	for i in range(n):
		b.append(list(map(int,list(input()))))
	p = []
	c = 0
	o = 0
	oo = m
	k = n
	kk = 0
	cc = 0
	for i in a:
		c+=sum(i)
	for i in b:
		cc+=sum(i)
	c = abs(cc-c)
	for i in range(n):
		pp = [0]*m
		for j in range(m):

			pp[j]=(int(b[i][j])-int(a[i][j]+3))%3
			if (int(b[i][j])-int(a[i][j]))%3!=0:
				o = max(o,j)
				oo=min(oo,j)
				kk=i
				k=min(i,k)
		p.append(pp)
	p = p[k:kk+1]
	for i in range(len(p)):
		p[i]=p[i][oo:o+1]
	g = True
	if p!=[]:
		h = len(p[0])

	for i in range(len(p)):
		for j in range(h):
			if p[i][j]!=p[-i-1][-j-1] and (p[i][j]==0 or p[-i-1][-j-1]==0):
				g= False
				break
		if not g:
			break

	if g and c%3==0:
		print("YES")
	else:
		print("NO")

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

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

Think step by step.

Response:
```plaintext
1
3 3
000
000
000
111
111
111
```

**Reasoning:**

The provided test input is designed to expose a potential bug in the given buggy code. Here's why this test case is challenging:

1. **Initial and Target Grids:**
   - The initial grid `a` is a 3x3 grid filled with zeros.
   - The target grid `b` is a 3x3 grid filled with ones.

2. **Transformation Requirement:**
   - To convert `a` into `b`, each cell in `a` needs to be incremented by 1 modulo 3. This can be achieved by applying the operation to the entire grid once, where all four corners are incremented by 1 and the other two corners are incremented by 2, effectively changing all zeros to ones.

3. **Buggy Code Analysis:**
   - The buggy code calculates the difference in sums between the initial and target grids (`c` and `cc`).
   - It then checks if the difference is divisible by 3, which is correct.
   - However, the code also checks the symmetry of the differences in the grid, which is not necessary for this specific transformation. The symmetry check might fail if the code incorrectly handles the indices or the conditions for symmetry.

4. **Expected Output:**
   - The correct output for this test case should be "YES" because it is possible to convert the initial grid `a` into the target grid `b` using the given operation.

This test case is small and simple, making it easy to verify the correctness of the solution. It focuses on a straightforward transformation that should be handled correctly by a well-implemented algorithm, but it can expose issues in the buggy code's symmetry checks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

Buggy Code:
# cook your dish here
for t in range(int(input())):
    n, m = map(int, input().split(" "))
    mat1 = []
    for i in range(n):
        l1 = input()
        mat1.append(l1)

    mat2  =[]
    for i in range(n):
        l1 = input()
        mat2.append(l1)

    temp1 = 0
    temp2 = 0
    ans = 1
    for i in range(m):
        temp1=0
        temp2=0
        for j in range(n):
            temp1+= int(mat1[j][i])
            temp2+= int(mat2[j][i])

        if temp1%3 != temp2%3:
            ans=0
    if ans:
        print("Yes")
    else:
        print("No")

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

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

Think step by step.

Response:
```plaintext
1
2 2
00
00
11
11
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

Buggy Code:
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class Aleksa {
    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int t = fs.nextInt();
        while (t-- != 0) {
            int n = fs.nextInt();
            int m = fs.nextInt();
            String[] a = new String[n];
            String[] b = new String[n];
            for(int i = 0 ;i<n ; i++){
                a[i] = fs.next();
            }
            for(int i = 0 ;i<n ; i++){
                b[i] = fs.next();
            }
            Boolean flag = true;
            int s1 = 0 , s2 = 0;
            for(int j = 0 ; j<m ; j++){
                s1 = 0; s2 = 0;
                for(int i = 0 ; i<n ; i++){
                    s1+= a[i].charAt(j)-'0';
                    s2+= b[i].charAt(j)-'0';
                }
                if (s1%3 != s2%3) {
                    flag = false;
                    break;
                }
            }
            if (flag) {
                out.println("Yes");
            }
            if (!flag) {
                out.println("No");
            }
        }
        out.close();
    }

    static class Key {
        long x, y, z;
        public Key(long x, long y, long z) {
            this.x = x;
            this.y = y;
            this.z = z;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Key key = (Key) o;
            return x == key.x && y == key.y && z == key.z;
        }

        @Override
        public int hashCode() {
            return Objects.hash(x, y, z);
        }
    }

    static class Pair {
        long x, y;
        public Pair(long x, long y) {
            this.x = x;
            this.y = y;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair pair = (Pair) o;
            return x == pair.x && y == pair.y;
        }

        @Override
        public int hashCode() {
            return Objects.hash(x, y);
        }
    }
    static class InnerPair<A, B> {
        private A first;
        private B second;

        public InnerPair(A first, B second) {
            this.first = first;
            this.second = second;
        }

        public A getFirst() {
            return first;
        }

        public B getSecond() {
            return second;
        }

        @Override
        public String toString() {
            return "(" + first + ", " + second + ")";
        }
    }
    static class OuterPair<A, B, C> {
        private InnerPair<A, B> innerPair;
        private C third;

        public OuterPair(A first, B second, C third) {
            this.innerPair = new InnerPair<>(first, second);
            this.third = third;
        }

        public InnerPair<A, B> getInnerPair() {
            return innerPair;
        }

        public C getThird() {
            return third;
        }

        @Override
        public String toString() {
            return "(" + innerPair + ", " + third + ")";
        }
    }

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

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

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

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

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the logic that checks the sum of the columns modulo 3. The code incorrectly assumes that if the sum of the columns modulo 3 is the same for both grids, then it is possible to convert grid `a` to grid `b`. However, this is not sufficient to guarantee that the transformation is possible.

### Steps to Generate the Test Input:
1. **Identify the Bug**: The bug lies in the assumption that the sum of the columns modulo 3 being equal is enough to determine if the transformation is possible. This is incorrect because the operation affects the corners of a subrectangle, and the modulo 3 check does not account for the specific changes in the grid.

2. **Create a Test Case**: We need a test case where the sum of the columns modulo 3 is the same for both grids, but the transformation is still impossible due to the specific nature of the operations.

3. **Example**: Consider a 3x3 grid where the sum of the columns modulo 3 is the same, but the specific values cannot be transformed into each other using the given operations.

### Test Input:
```plaintext
1
3 3
000
000
000
111
222
000
```

### Explanation:
- **Grid `a`**:
  ```
  000
  000
  000
  ```

- **Grid `b`**:
  ```
  111
  222
  000
  ```

- **Column Sums**:
  - For grid `a`: All columns sum to 0.
  - For grid `b`: All columns sum to 3, which is 0 modulo 3.

- **Transformation Impossibility**:
  - The top-left corner of the subrectangle in grid `a` is 0, and it needs to be 1 in grid `b`.
  - The top-right corner of the subrectangle in grid `a` is 0, and it needs to be 1 in grid `b`.
  - The bottom-left corner of the subrectangle in grid `a` is 0, and it needs to be 1 in grid `b`.
  - The bottom-right corner of the subrectangle in grid `b` is 0, and it needs to be 1 in grid `b`.

  However, the operation can only change the values in pairs, and it is impossible to achieve the required transformation using the given operations.

This test case will expose the bug in the code because the code will incorrectly output "Yes" even though the transformation is impossible.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

Buggy Code:
import java.util.Scanner;

public class corners {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t=scan.nextInt();
        while(t-->0)
        {
            int n=scan.nextInt();
            int m=scan.nextInt();
            String a[]=new String[n];
            String b[]=new String[n];
            for(int i=0;i<n;i++)
            {
                a[i]=scan.next();
            }
            for(int i=0;i<n;i++)
            {
                b[i]=scan.next();
            }
            int flag=0;
            for(int i=0;i<m;i++)
            {
                int suma=0;
                int sumb=0;
                for(int j=0;j<n;j++)
                {
                    suma+=(int)(a[j].charAt(i)-48);
                    sumb+=(int)(b[j].charAt(i)-48);
                }
                if(suma%3!=sumb%3)
                {
                    flag=1;
                    System.out.println("NO");
                    break;
                }
            }
            if(flag==0)
            {
                System.out.println("YES");
            }
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on scenarios where the code's logic fails. The buggy code checks if the sum of each column in grid `a` is congruent to the sum of the corresponding column in grid `b` modulo 3. However, this is not sufficient to determine if the transformation is possible. The correct approach would involve checking the parity of the number of 1s and 2s in the grid, as the operations change the values in specific ways.

### Step-by-Step Reasoning:

1. **Identify the Bug**: The bug is in the logic that only checks the column sums modulo 3. This is insufficient because the operations affect the values in a more complex way, particularly in terms of the number of 1s and 2s.

2. **Generate a Test Case**: We need a test case where the column sums modulo 3 are the same, but the transformation is impossible due to the parity of 1s and 2s.

3. **Example Test Case**:
   - **Grid `a`**: A 3x3 grid with all zeros.
   - **Grid `b`**: A 3x3 grid with a specific pattern that cannot be achieved by the given operations.

### Chosen Test Case:
- **Grid `a`**:
  ```
  000
  000
  000
  ```
- **Grid `b`**:
  ```
  012
  102
  220
  ```

### Explanation:
- **Grid `a`** is all zeros.
- **Grid `b`** has a pattern that cannot be achieved by the operations described. Specifically, the transformation from all zeros to the pattern in `b` would require an odd number of operations in some subrectangles, which is not possible given the constraints of the operations.

### Test Input:
```plaintext
1
3 3
000
000
000
012
102
220
```

This test case should expose the bug in the code, as the column sums modulo 3 are the same, but the transformation is impossible.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

Buggy Code:
import math # Leeisateam
import itertools
import random
from heapq import *
from collections import deque # Usage: d=deque([])
import sys
def input():return sys.stdin.readline().strip() # fast IO
#sys.setrecursionlimit(100000) # Will MLE PyPy but not Python. 100000 -> 128MB
debug=1

T=int(input())
for TC in range(T):
    n,m=map(int,input().split())
    a=[]
    b=[]
    for i in range(n):
        a.append(list(map(int,list(input()))))
    for i in range(n):
        b.append(list(map(int,list(input()))))

    for i in range(n-1):
        for j in range(m-1):
            c=(b[i][j] - a[i][j]+3)%3
            a[i][j] += c
            a[i+1][j+1] += c
            a[i][j+1] += 3-c
            a[i+1][j] += 3-c

            a[i][j]%=3
            a[i][j+1]%=3
            a[i+1][j]%=3
            a[i+1][j+1]%=3

    ans="YES"
    for i in range(n):
        for j in range(n):
            if a[i][j] != b[i][j]:
                ans="NO"

    #print(*a)
    #print(*b)
    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
000
000
000
012
120
201
```

**Explanation:**

1. **Grid Size and Content:**
   - The grid is 3x3, which is small enough to be manageable but large enough to show the issue.
   - The initial grid `a` is filled with zeros.
   - The target grid `b` has a specific pattern where each corner of the grid is different.

2. **Operation Explanation:**
   - The operation changes the values of the corners of a subrectangle. For a 3x3 grid, the only subrectangle that can be chosen is the entire grid itself.
   - When the entire grid is chosen, the corners (top-left, top-right, bottom-left, bottom-right) are the only positions affected.

3. **Expected Outcome:**
   - The target grid `b` has the values:
     ```
     0 1 2
     1 2 0
     2 0 1
     ```
   - To achieve this, the corners must be transformed as follows:
     - Top-left: 0 -> 0 (unchanged)
     - Top-right: 0 -> 1 (incremented by 1)
     - Bottom-left: 0 -> 2 (incremented by 2)
     - Bottom-right: 0 -> 1 (incremented by 1)

4. **Challenge:**
   - The buggy code might not correctly handle the transformation of the corners, especially when the values need to be incremented by different amounts.
   - The specific pattern in `b` ensures that the buggy code will likely fail to correctly transform `a` into `b`.

This test case is designed to expose the bug in the code by focusing on the corner transformations, which are critical in this problem.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

Buggy Code:
import sys
from collections import defaultdict
input=sys.stdin.readline
for _ in range(int(input())):
    n,m=map(int,input().split())
    a = []
    for _ in range(n):
        row=tuple(input().strip())
        a.append(row)

    b = []
    for _ in range(n):
        row=tuple(input().strip())
        b.append(row)

    diff_array = []

    for i in range(n):
        row = []
        for j in range(m):
            element_diff = int(b[i][j]) - int(a[i][j])
            if element_diff == -1:
                row.append(2)
            elif element_diff == -2:
                row.append(1)
            else:
                row.append(element_diff)

        diff_array.append(row)

    all_rows_sums = [sum(x)%3 for x in diff_array]
    all_cols_sums = []

    for j in range(m):
        acc = 0
        for i in range(n):
            acc += diff_array[i][j]

        all_cols_sums.append(acc%3)

    # print(all_rows_sums)
    # print(all_cols_sums)
    # print()

    set_of_both = set(all_rows_sums+all_cols_sums)
    if len(set_of_both)==1:
        print("YES")
    else:
        print("NO")

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

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

Think step by step.

Response:
```plaintext
1
3 3
000
000
000
121
212
121
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

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

using namespace std;

void solve(){
  int n, m;
  cin >> n >> m;
  string a[n], b[n];
  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++){
      for (int j = 0; j < m; j++){
          cin >> b[i][j];
      }
  }
  for (int i = 0; i < n; i++){
      int cnt_a = 0, cnt_b = 0;
      for (int j = 0; j < m; j++){
          cnt_a = (cnt_a + a[i][j]) % 3;
          cnt_b = (cnt_b + b[i][j]) % 3;
      }
      if (cnt_a != cnt_b){
          cout << "NO" << endl;
          return;
      }
  }
  for (int i = 0; i < m; i++){
      int cnt_a = 0, cnt_b = 0;
      for (int j = 0; j < n; j++){
          cnt_a = (cnt_a + a[j][i]) % 3;
          cnt_b = (cnt_b + b[j][i]) % 3;
      }
      if (cnt_a != cnt_b){
          cout << "NO" << endl;
          return;
      }
  }
  cout << "YES" << endl;
}
int main(){
    int T;
    cin >> T;
    while(T--){
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
2 2
00
00
11
11
2 2
00
00
22
22
2 2
00
00
12
21
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
/*
3 3

012
012
012

010
111
011

row = [2 0 1]
col = [2 0 1]
row3 = 2, 1
col2 = 2

001
111
111

111
111
111

row = [0 0 0]
col = [1 1 1]

*/
int main() {
	int t;
	cin>>t;
	while(t--) {
	    int n,m;
	    cin>>n>>m;
	    vector<string> a(n),b(n);
	    for(int i=0;i<n;i++) cin>>a[i];
	    for(int i=0;i<n;i++) cin>>b[i];

	    vector<int> row(n,0), col(m,0);
	    for(int i=n-1;i>=0;i--) {
	        for(int j=m-1;j>=0;j--) {
	            if(a[i][j] != b[i][j]) {
	                int n1 = a[i][j]-'0';
	                int n2 = b[i][j]-'0';

                    int pos = (3 - n2 + n1)%3;
                    row[i]=(row[i]+pos)%3;
                    col[j]=(col[j]+pos)%3;

	            }
	        }
	    }

	    int res=1;
	    for(int i=0;i<n;i++) {
	        if(row[i] != 0 || col[i]!=0) res=0;
	    }
	    if(res==1) cout<<"YES"<<endl;
	    else cout<<"NO"<<endl;
	}
}

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

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

Think step by step.

Response:
```plaintext
2
3 3
000
000
000
111
111
111
3 3
012
012
012
010
111
011
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

Buggy Code:
import java.util.*;
import java.io.*;
public class Main{
	static FastScanner fs = new FastScanner();
	static PrintWriter out = new PrintWriter(System.out);
	static InputStream is;
	public static void main(String args[]){
		int t =fs.nextInt();
	   while(t-->0){
	   	solve();
	   	out.flush();
	   }
	}

    static void solve(){
	  int n  = fs.nextInt();
	  int m = fs.nextInt();
	  int a[][] = new int[n][m];
	  int b[][] = new int[n][m];

	  for(int i  = 0;i<n;i++) {
		  String s = fs.next();

		  for(int j= 0;j<m;j++) {
			  a[i][j] = s.charAt(j)-'0';
		  }
	  }

	  for(int i  = 0;i<n;i++) {
		  String s = fs.next();

		  for(int j= 0;j<m;j++) {
			  b[i][j] = s.charAt(j)-'0';
		  }
	  }

	  int f[] = new int[3];
	  for(int i = 0;i<n-1;i++) {
		  for(int j = 0;j<m-1;j++) {
			  int diff =b[i][j] - a[i][j];
			  if(diff == 0)continue;

			  if(diff<0)diff+=3;

			  a[i][j] += diff;
			  a[i+1][j+1] += diff;
			  a[i+1][j] += (3-diff);
			  a[i][j+1] += (3-diff);

		  }
	  }

	  for(int i = 0;i<n;i++) {
		  for(int j = 0;j<m;j++) {
			  if((a[i][j])%3!=(b[i][j])%3) {
				  out.println("No");
				  return;
			  }
		  }
	  }

	  out.println("Yes");
	}

	static void sieve(int n,int p[]){
        for(int i=2;i<=n;i++){
            if(p[i] == 0)continue;
            for(long j=(long)i*(long) i;j<=n;j+=(long)i){
                p[(int)j]=0;
            }
        }
    }

	static int gcd(int a,int b){
        if(a == 0)return b;
        return gcd(b%a,a);
    }

	 public static long p(int n,long base){
	        long ans = 1L;
	        long mod = (long) Math.pow(10,9)+7L;

	      while(n>0){
	            if(n%2 == 1){
	                ans =  (ans%mod*base)%mod;
	                n-= 1;
	            }else{
	                base = (base%mod*base)%mod;
	                n /= 2;
	            }

	        }
	        return ans;
	    }

	static void clos(double d) {

	}
}

class Pair{
    int u;
    int v;

    public Pair(int u, int v) {
        this.u = Math.min(u,v);
        this.v = Math.max(u,v);
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof Pair)) return false;
        Pair pair = (Pair) o;
        return u == pair.u && v == pair.v;
    }

    @Override
    public int hashCode() {
        return Objects.hash(u, v);
    }
}

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

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i <n; i++)
                a[i] = nextInt();
            return a;
        }
        long[] readArrayL(int n) {
        	 long[] a = new long[n];
             for (int i = 0; i < n; i++)
                 a[i] = nextLong();
             return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        Double nextDouble() {
        	return Double.parseDouble(next());
        }

  }

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases where the code might fail. The bug in the code lies in the way it handles the transformation of the grid. Specifically, the code might not correctly account for all possible transformations, especially in cases where the grid has a specific pattern or configuration that makes it impossible to transform from `a` to `b`.

### Step-by-Step Reasoning:

1. **Identify the Bug**:
   - The code attempts to transform the grid by applying the operation to subrectangles and checking if the final grid matches the target grid.
   - The bug might be in the way the code handles the transformation, particularly in cases where the grid has a specific pattern that cannot be achieved by the allowed operations.

2. **Consider Edge Cases**:
   - **Single Element Subrectangles**: The code does not handle single element subrectangles correctly, as the operation requires a subrectangle of at least 2x2.
   - **Corner Cases**: The code might fail when the grid has a specific pattern that cannot be transformed by the allowed operations.

3. **Generate a Test Case**:
   - We need a grid that is small enough to fit the output limit but complex enough to expose the bug.
   - A 3x3 grid is a good choice because it is small and can be easily analyzed.

### Challenging Test Input:

Let's create a 3x3 grid `a` and a 3x3 grid `b` such that it is impossible to transform `a` into `b` using the given operations.

```plaintext
1
3 3
000
000
000
222
222
222
```

### Explanation:

- **Grid `a`**:
  - All elements are 0.

- **Grid `b`**:
  - All elements are 2.

### Why This Test Case is Challenging:

- **Transformation Analysis**:
  - To transform a 0 to a 2, the operation must be applied in such a way that the corners of the subrectangle are incremented by 2 modulo 3.
  - However, to achieve a uniform grid of 2s from a uniform grid of 0s, every element in `a` must be incremented by 2 modulo 3.
  - The operation affects only the corners of the subrectangle, and it is impossible to uniformly increment every element in a 3x3 grid from 0 to 2 using the allowed operations.

This test case should expose the bug in the code, as the code will incorrectly determine that the transformation is possible, leading to an incorrect output of "Yes" instead of "No".


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Corner Twist
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two grids of numbers $a$ and $b$, with $n$ rows and $m$ columns. All the values in the grid are $0$, $1$ or $2$.

You can perform the following operation on $a$ any number of times:

  * Pick any subrectangle in the grid with length and width $\ge 2$. You are allowed to choose the entire grid as a subrectangle.    * The subrectangle has four corners. Take any pair of diagonally opposite corners of the chosen subrectangle and add $1$ to their values modulo $3$.    * For the pair of corners not picked, add $2$ to their values modulo $3$.

Note that the operation only changes the values of the corners of the picked subrectangle.

Is it possible to convert the grid $a$ into grid $b$ by applying the above operation any number of times (possibly zero)?

Input: The first line contains an integer $t$, the number of testcases ($1 \le t \le 250$).

For each testcase:

The first line contains two integers $n$ and $m$, the number of rows and columns in the grid ($2 \le n,m \le 500$).

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $a_{i,j}$.

Each of the next n lines contain m characters — the $j$-th character of the $i$-th line represents $b_{i,j}$ ($0 \le a_{i,j}, b_{i,j} \le 2$).

It is guaranteed that the sum of $n$ over all test cases and the sum of $m$ over all test cases do not exceed $500$.

Output: For each test case print "YES" (without quotes) if it is possible to convert grid $a$ into grid $b$ and "NO" (without quotes) otherwise.

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

Examples:
input:
7
3 3
000
000
000
111
111
111
4 4
0000
0000
0000
0000
2100
1200
0012
0021
4 4
1020
1200
1210
0000
0000
1200
2200
0000
3 3
012
012
012
010
111
011
8 8
00000000
00000000
00000000
00000000
00000000
00000000
00000000
10000000
00000000
01200000
02010000
00102000
00020100
00001020
00000210
10000000
2 7
0000000
0000000
2220111
0111222
2 7
0000000
0100010
2220111
1210202
output:
YES
YES
YES
NO
YES
NO
YES

Note: In the first testcase, grid $a$ can be converted into $b$ in the following manner:

$\begin{matrix}\fbox{0} & 0 & \fbox{0}\\\ 0 & 0 & 0\\\ \fbox{0} & 0 & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ 0 & \fbox{0} & \fbox{0}\\\ 2 & \fbox{0} & \fbox{1}\end{matrix} \Rightarrow \begin{matrix}1 & 0 & 2\\\ \fbox{0} & \fbox{1} & 2\\\ \fbox{2} & \fbox{2} & 2\end{matrix} \Rightarrow \begin{matrix}1 & \fbox{0} & \fbox{2}\\\ 1 & 0 & 2\\\ 1 & \fbox{0} & \fbox{2}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & \fbox{0} & \fbox{2}\\\ 1 & \fbox{2} & \fbox{0}\end{matrix} \Rightarrow \begin{matrix}1 & 1 & 1\\\ 1 & 1 & 1\\\ 1 & 1 & 1\end{matrix}$

Here, in each operation, the top-right and bottom-left corners highlighted by a box are incremented by $2$ modulo $3$, while the top-left and bottom-right corners are incremented by $1$ modulo $3$.

In the fourth testcase, it can be proven that it is not possible to convert grid $a$ into grid $b$ using the above-mentioned operations any number of times.

Buggy Code:
import java.util.Scanner;

public class Main {

    public static void readMatrix(Scanner scanner, int[][] matrix, int rows, int cols) {
        for (int i = 0; i < rows; i++) {
            String line = scanner.next();
            for (int j = 0; j < cols; j++) {
                if (line.charAt(j) == '0') {
                    matrix[i][j] = 0;
                } else if (line.charAt(j) == '1') {
                    matrix[i][j] = 1;
                } else if (line.charAt(j) == '2') {
                    matrix[i][j] = 2;
                }
            }
        }
    }

    public static boolean solve(int[][] matrixA, int[][] matrixB, int rows, int cols) {
        boolean flag = true;
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (matrixA[i][j] != matrixB[i][j]) {
                    int value = matrixA[i][j];
                    if (i + 1 < rows && j + 1 < cols) {
                        if ((value + 2) % 3 == matrixB[i][j]) {
                            matrixA[i + 1][j + 1] = (matrixA[i + 1][j + 1] + 2) % 3;
                            matrixA[i][j + 1] = (matrixA[i][j + 1] + 1) % 3;
                            matrixA[i + 1][j] = (matrixA[i + 1][j] + 1) % 3;
                            matrixA[i][j] = matrixB[i][j];
                        } else if ((value + 1) % 3 == matrixB[i][j]) {
                            matrixA[i + 1][j + 1] = (matrixA[i + 1][j + 1] + 1) % 3;
                            matrixA[i][j + 1] = (matrixA[i][j + 1] + 2) % 3;
                            matrixA[i + 1][j] = (matrixA[i + 1][j] + 2) % 3;
                            matrixA[i][j] = matrixB[i][j];
                        } else {
                            flag = false;
                            break;
                        }
                    } else {
                        flag = false;
                        break;
                    }
                }
            }
            if (!flag) {
                break;
            }
        }
        return flag;
    }

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

        while (testCases-- > 0) {
            int rows = scanner.nextInt();
            int cols = scanner.nextInt();

            int[][] matrixA = new int[100][100];
            int[][] matrixB = new int[100][100];

            readMatrix(scanner, matrixA, rows, cols);
            readMatrix(scanner, matrixB, rows, cols);

            if (solve(matrixA, matrixB, rows, cols)) {
                System.out.println("YES");
            } else {
                System.out.println("NO");
            }
        }
        scanner.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3 3
000
000
000
222
222
222
```

### Explanation:
This test case is designed to expose the issue in the buggy code. Here’s why:

1. **Initial Grids**:
   - Grid `a` is all zeros:
     ```
     000
     000
     000
     ```
   - Grid `b` is all twos:
     ```
     222
     222
     222
     ```

2. **Expected Behavior**:
   - To convert `a` to `b`, each cell in `a` needs to be incremented by 2 modulo 3. However, the operation can only affect the corners of a subrectangle, and the changes are not uniformly applied to all cells.

3. **Buggy Code Issue**:
   - The code attempts to convert each cell individually by checking if it can be incremented by 1 or 2 to match the target value in `b`.
   - However, the code does not consider the global constraints of the problem. It tries to convert each cell in isolation, which is not feasible with the given operation constraints.

4. **Why This Test Case Fails**:
   - The code will attempt to convert the first cell (0,0) from 0 to 2. It will try to pick a subrectangle that includes (0,0) and (1,1) and apply the operation. However, this will also affect the other two corners (0,1) and (1,0), which are not initially set to the correct values.
   - The code will continue to try to convert each cell individually, but it will eventually fail to convert the entire grid to the desired state because the operation affects multiple cells simultaneously, and the code does not handle this correctly.

This test case should cause the buggy code to output "NO" because it cannot correctly handle the transformation of the entire grid.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

Input: The first line contains an integer $t$ ($1 \le t \le 100$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
#include <iostream>
#include <iomanip>
#include <string>
#include <cstring>
#include <vector>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <numeric>
#include <map>
#include <list>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <limits.h>
#include <stack>
using namespace std;

#define forn(i, n) for (int i = 0; i < n; i++)
#define form2n(i, m, n) for (long long i = m; i < n; i++)
#define ll long long
#define pii pair<ll, ll>
#define vi vector<int>
#define vll vector<long long>
#define vpii vector<pii>
#define yesno(b) if (b) cout << "YES\n"; else cout << "NO\n";
#define lcm(a, b) (a * b) / __gcd(a, b)
#define INF 1000

ll func(const vpii& arr, const ll& n, const ll& k, ll ans, ll points, ll mask, map<ll, ll>& m, ll index)
{
    mask = mask ^ (1 << index);
    if (m.find(mask) == m.end())
    {
        if (arr[index].first + arr[index].second + points < k)
        {
            points += (arr[index].first + arr[index].second);
            ans += (arr[index].first * arr[index].second);
            ll a = LONG_MAX;
            forn(i, n)
            {
                if (mask & (1 << i))
                    a = min(a, func(arr, n, k, ans, points, mask, m, i));
            }
            m[mask] = a;
        }
        else if (arr[index].first + arr[index].second + points == k ||
                arr[index].first + arr[index].second + points == k + 1)
        {
            m[mask] = ans + (arr[index].first * arr[index].second);
        }
        else
        {
            // int p = min(arr[index].second - 1, k - points);
            // ans += arr[index].first * p;
            // points += p;
            // if (points == k)
            //     m[mask] = ans;
            // else
            //     m[mask] = ans + (k - points);

            ll rem = k - points;
            ll a = LONG_MAX;
            for (ll i = min(rem, arr[index].first - 1); i >= 0; i--)
            {
                ll j = rem - i;
                if (j >= arr[index].second)
                    break;
                a = min(a, arr[index].second * i + arr[index].first * j - i * j);
            }
            m[mask] = a + ans;
        }
    }
    return m[mask];
}

int main(void)
{
    ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

    int t = 1;
    cin >> t;
    while (t--)
    {
        ll n, k;
        cin >> n >> k;
        vpii arr;
        ll h, w;
        ll total = 0;
        forn(i, n)
        {
            cin >> h >> w;
            ll mn = min(h, w);
            ll mx = max(h, w);
            arr.push_back({mn, mx});
            total += (h + w);
        }
        if (total < k)
            cout << -1 << endl;
        else
        {
            map<ll, ll> m;
            ll ans = LONG_MAX;
            forn(i, n)
                ans = min(ans, func(arr, n, k, 0, 0, (1 << n) - 1, m, i));
            cout << ans << endl;
        }

        // sort(arr.begin(), arr.end());
        // ll ans = 0;
        // int points = 0;
        // int i = 0;
        // bool b = true;
        // while (points < k)
        // {
        //     if (i == n)
        //     {
        //         b = false;
        //         break;
        //     }
        //     if (arr[i] == pii{1, 1})
        //     {
        //         ans += 1;
        //         points += 2;
        //         i++;
        //         continue;
        //     }
        //     int p = min(arr[i].second - 1, k - points);
        //     ans += arr[i].first * p;
        //     points += p;
        //     if (points == k)
        //         break;
        //     p = min(arr[i].first, k - points);
        //     ans += p;
        //     points += (p + 1);
        //     i++;
        // }
        // if (b)
        //     cout << ans << endl;
        // else
        //     cout << -1 << endl;
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

Input: The first line contains an integer $t$ ($1 \le t \le 100$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
const int MAXN =1e3+10;
const int inf=1e9;

struct mt
{
    int v,w;//价值，费用
    mt(int V,int W){v=V,w=W;}
};
vector<mt> a[MAXN];
int f[MAXN][MAXN];

void solve()
{
    int n,k;
    cin>>n>>k;
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        a[i].clear();
        int x,y;
        cin>>x>>y;
        sum+=x*y;
        int tot=0,count=0;
        while(x>0 && y>0)
        {
            //cout<<x<<" "<<y<<endl;
            if(x==1 && y==1)
            {
                tot+=1;
                count+=2;
                a[i].push_back(mt(count,tot));
                x--;
                y--;
            }
            if(x<y)swap(x,y);
            if(x>=y && x>0 && y>0)
            {
                tot+=y;
                count++;
                a[i].push_back(mt(count,tot));
                x--;
            }
        }
    }
    //cout<<"OK"<<endl;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=sum;j++)f[i][j]=0;
    }
    int ans=inf;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=sum;j++)
        {
            for(auto x:a[i])
            {
                if(j>=x.w)
                {
                    f[i][j]=max(f[i][j],max(f[i-1][j-x.w]+x.v,f[i-1][j]));
                }
            }
            if(f[i][j]>=k)
            {
                ans=min(ans,j);
            }
        }
    }
    if(ans==inf)cout<<-1<<"\n";
    else cout<<ans<<"\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);
    int t=1;
    cin>>t;
    while(t--)solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

Input: The first line contains an integer $t$ ($1 \le t \le 100$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
// Problem: F. Color Rows and Columns
// Contest: Codeforces - Codeforces Round 966 (Div. 3)
// URL: https://codeforces.com/contest/2000/problem/F
// Memory Limit: 256 MB
// Time Limit: 3000 ms
//
// Powered by CP Editor (https://cpeditor.org)

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

typedef long long ll;
typedef long double ld;
typedef pair<ll , ll> PII ;
typedef vector<bool> vb ;
typedef vector<ll> vi;
typedef vector<PII> vpi;
typedef vector<vector<ll>> vvi;
typedef vector<vector<PII>> vvpi;
typedef vector<char> vc ;

const double EBS = 1e-9;
const double pi = 3.14159265358979 ;

#define Log(x) (31^__builtin_clz(x))
#define logll(x) (63^__builtin_clzll(x))

#define rand __builtin_ia32_rdtsc()

#define popcount(i) __builtin_popcount(i)
#define popcountll(i) __builtin_popcountll(i)

#define mp make_pair

#define Time cerr << "Time Taken: " << (float)clock() / CLOCKS_PER_SEC << " Secs"  << "\n";

#define testCase ll t; cin >> t; while (t--)
#define fast  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define rep(f, s, i) for (ll i = f; (int) i < s; i++)
#define getunique(v) {sort(v.begin(), v.end()); v.erase(unique(v.begin(), v.end()), v.end());}
#define calunique(v)  disance(v.begin(),unique(v.begin(),v.end()));
#define el cout << endl
#define pb push_back
#define no cout << "NO\n"
#define yes cout << "YES\n"
#define all(v) v.begin(), v.end()
#define INF (ll) 1e9
#define INFLL (ll)1e18
#define debug cout << "\n___________________________________\n"
#define int ll

#define Left  ((p << 1) + 1)
#define Right ((p << 1) + 2)
void solve(){
	int n , k ; cin >> n >> k ;
	vpi a(n);
	int all = 0 ;
	rep(0 , n , i){
		cin >> a[i].first >> a[i].second;
		all += a[i].first + a[i].second;
		if(a[i].first > a[i].second)swap(a[i].first , a[i].second);
	}
	if(all < k){
		cout << "-1\n";
		return ;
	}
	int score = 0 ;
	int sum = 0;
	multiset<PII> s , temp ;
	rep(0 , n , i){
		s.insert(a[i]);
	}
	for(auto [x , y] : s){
		if(x + y + score <= k){
			sum += x * y ;
			score += x + y ;
		}else{
			temp.insert(mp(x , y));
		}
	}
	swap(s , temp);
	while(score < k){
		auto [x , y] = *s.begin();

		s.erase(s.begin());
		sum += x ;
		y--;
		score++;
		if(x == 0 && y == 0){
			continue;
		}
		if(x > y) swap(x , y);
		s.insert(mp(x , y));
	}
	cout << sum << endl;

}

int32_t main(){
	fast ;
	testCase{
		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 10
10 1
10 1
10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Color Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have $n$ rectangles, the $i$-th of which has a width of $a_i$ and a height of $b_i$.

You can perform the following operation an unlimited number of times: choose a rectangle and a cell in it, and then color it.

Each time you completely color any row or column, you earn $1$ point. Your task is to score at least $k$ points with as few operations as possible.

Suppose you have a rectangle with a width of $6$ and a height of $3$. You can score $4$ points by coloring all the cells in any $4$ columns, thus performing $12$ operations.

Input: The first line contains an integer $t$ ($1 \le t \le 100$) — the number of test cases. The following are the descriptions of the test cases.

The first line of each test case description contains two integers $n$ and $k$ ($1 \le n \le 1000, 1 \le k \le 100$) — the number of rectangles in the case and the required number of points.

The next $n$ lines contain the descriptions of the rectangles. The $i$-th line contains two integers $a_i$ and $b_i$ ($1 \le a_i, b_i \le 100$) — the width and height of the $i$-th rectangle.

It is guaranteed that the sum of the values of $n$ across all test cases does not exceed $1000$.

Output: For each test case, output a single integer — the minimum number of operations required to score at least $k$ points. If it is impossible to score at least $k$ points, output -1.

Examples:
input:
7
1 4
6 3
1 5
4 4
5 10
1 1
1 1
1 1
1 1
1 1
2 100
1 2
5 6
3 11
2 2
3 3
4 4
3 25
9 2
4 3
8 10
4 18
5 4
8 5
8 3
6 2
output:
12
14
5
-1
17
80
35

Note:

Buggy Code:
size = int(input())

for i in range(size):
    n, k = map(int, input().split(' '))
    recs = []
    k_bound = 0
    for j in range(n):
        a,b = map(int, input().split(' '))
        recs.append((a, b))
        k_bound += (a + b)

    if k > k_bound:
        print('-1')
        continue

    # initialise opt[n][k] as subproblem
    opt = [[float('inf') for _ in range(k_bound + 1)] for _ in range(n + 1)]

    # pts and ops are on each recs for once
    for i in range(1, n + 1):
        pts = 0
        ops = 0
        while True:
            if pts > k_bound:
                break
            w, h = recs[i - 1]
            if w == 1 and h == 1:
                pts += 2
                ops += 1
                recs[i - 1] = (0, 0)
            elif w == 0 or h == 0:
                # current rec is fully colored, move to next rec
                break
            else:
                pts += 1
                ops += min(w, h)
                # update the rec after operations
                recs[i - 1] = (min(w, h), max(w, h) - 1)

            # update opt[i][j] if current operation is cheaper
            if pts <= k_bound:
                opt[i][pts] = min(opt[i - 1][pts], ops)
                # if opt[i][pts] != float('inf'):
                #     print(f" 1- update opt[{i}][{pts}] to be {opt[i][pts]}")
            for j in range(1, k_bound + 1):
                if j + pts <= k_bound:
                    opt[i][j + pts] = min(opt[i][j + pts], opt[i - 1][j] + ops)
                    # if opt[i][j + pts] != float('inf'):
                    #     print(f"2- update opt[{i}][{j + pts}] to be {opt[i][j + pts]}")

    ans = min(opt[-1][k:])
    print(ans) if ans != float('inf') else print(-1)

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

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

Think step by step.

Response:
```plaintext
3
1 2
2 2
1 3
2 2
3 6
2 3
3 2
4 1
```