**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: King Keykhosrow's Mystery
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a tale about the wise King Keykhosrow who owned a grand treasury filled with treasures from across the Persian Empire. However, to prevent theft and ensure the safety of his wealth, King Keykhosrow's vault was sealed with a magical lock that could only be opened by solving a riddle.

The riddle involves two sacred numbers $$$a$$$ and $$$b$$$. To unlock the vault, the challenger must determine the smallest key number $$$m$$$ that satisfies two conditions:

- $$$m$$$ must be greater than or equal to at least one of $$$a$$$ and $$$b$$$.
- The remainder when $$$m$$$ is divided by $$$a$$$ must be equal to the remainder when $$$m$$$ is divided by $$$b$$$.

Only by finding the smallest correct value of $$$m$$$ can one unlock the vault and access the legendary treasures!

Input: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$), the number of test cases.

Each test case consists of a single line containing two integers $$$a$$$ and $$$b$$$ ($$$1 \leq a, b \leq 1000$$$).

Output: For each test case, print the smallest integer $$$m$$$ that satisfies the conditions above.

Examples:
input:
2
4 6
472 896
output:
12
52864

Note: In the first test case, you can see that:

- $$$4 \bmod 4 = 0$$$ but $$$4 \bmod 6 = 4$$$
- $$$5 \bmod 4 = 1$$$ but $$$5 \bmod 6 = 5$$$
- $$$6 \bmod 4 = 2$$$ but $$$6 \bmod 6 = 0$$$
- $$$7 \bmod 4 = 3$$$ but $$$7 \bmod 6 = 1$$$
- $$$8 \bmod 4 = 0$$$ but $$$8 \bmod 6 = 2$$$
- $$$9 \bmod 4 = 1$$$ but $$$9 \bmod 6 = 3$$$
- $$$10 \bmod 4 = 2$$$ but $$$10 \bmod 6 = 4$$$
- $$$11 \bmod 4 = 3$$$ but $$$11 \bmod 6 = 5$$$

Buggy Code:
def main():
    t = int(input())
    for _ in range(t):
        a, b = map(int, input().split())
        m = max(a, b)
        while True:
            if m % a == m % b:
                print(m)
                break
            m += 1

main()

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the code might fail. The code uses a simple incremental approach to find the smallest \( m \) that satisfies the conditions. However, this approach can be inefficient and might fail for certain edge cases.

### Edge Cases to Consider:
1. **When \( a \) and \( b \) are equal**: In this case, the smallest \( m \) should be \( a \) (or \( b \)).
2. **When \( a \) and \( b \) are coprime**: The smallest \( m \) will be \( a \times b \).
3. **When \( a \) is a multiple of \( b \) or vice versa**: The smallest \( m \) should be the larger of the two.
4. **When \( a \) and \( b \) are very large**: This can cause the incremental approach to take a long time.

### Test Input:
Let's create a test input that includes these edge cases:

```plaintext
4
1 1
2 3
4 8
999 1000
```

### Explanation:
1. **Test Case 1: \( a = 1 \), \( b = 1 \)**:
   - The smallest \( m \) should be 1.
2. **Test Case 2: \( a = 2 \), \( b = 3 \)**:
   - The smallest \( m \) should be 6.
3. **Test Case 3: \( a = 4 \), \( b = 8 \)**:
   - The smallest \( m \) should be 8.
4. **Test Case 4: \( a = 999 \), \( b = 1000 \)**:
   - The smallest \( m \) should be 999000.

These test cases cover a range of scenarios, including equal numbers, coprime numbers, and large numbers, which should help expose any issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: King Keykhosrow's Mystery
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a tale about the wise King Keykhosrow who owned a grand treasury filled with treasures from across the Persian Empire. However, to prevent theft and ensure the safety of his wealth, King Keykhosrow's vault was sealed with a magical lock that could only be opened by solving a riddle.

The riddle involves two sacred numbers $$$a$$$ and $$$b$$$. To unlock the vault, the challenger must determine the smallest key number $$$m$$$ that satisfies two conditions:

- $$$m$$$ must be greater than or equal to at least one of $$$a$$$ and $$$b$$$.
- The remainder when $$$m$$$ is divided by $$$a$$$ must be equal to the remainder when $$$m$$$ is divided by $$$b$$$.

Only by finding the smallest correct value of $$$m$$$ can one unlock the vault and access the legendary treasures!

Input: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$), the number of test cases.

Each test case consists of a single line containing two integers $$$a$$$ and $$$b$$$ ($$$1 \leq a, b \leq 1000$$$).

Output: For each test case, print the smallest integer $$$m$$$ that satisfies the conditions above.

Examples:
input:
2
4 6
472 896
output:
12
52864

Note: In the first test case, you can see that:

- $$$4 \bmod 4 = 0$$$ but $$$4 \bmod 6 = 4$$$
- $$$5 \bmod 4 = 1$$$ but $$$5 \bmod 6 = 5$$$
- $$$6 \bmod 4 = 2$$$ but $$$6 \bmod 6 = 0$$$
- $$$7 \bmod 4 = 3$$$ but $$$7 \bmod 6 = 1$$$
- $$$8 \bmod 4 = 0$$$ but $$$8 \bmod 6 = 2$$$
- $$$9 \bmod 4 = 1$$$ but $$$9 \bmod 6 = 3$$$
- $$$10 \bmod 4 = 2$$$ but $$$10 \bmod 6 = 4$$$
- $$$11 \bmod 4 = 3$$$ but $$$11 \bmod 6 = 5$$$

Buggy Code:
import java.lang.*;
import java.util.*;
public class KingKeykhosrowMystery
{
          public static void main(String args[])
          {
                    Scanner sc=new Scanner(System.in);
                    int t=sc.nextInt();
                    int c=0;
                    while(t>0)
                    {
                              int a=sc.nextInt();
                              int b=sc.nextInt();
                              int m=Math.max(a,b);
                              for(int i=m;i<100000;i++)
                              {
                                        if((i%a) == (i%b))
                                           {
                                                     c=i;
                                                     break;
                                           }
                              }
                              System.out.println(c);
                              t--;
                    }
          }
}

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

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

Think step by step.

Response:
```plaintext
3
1 1
1000 1000
999 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: King Keykhosrow's Mystery
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a tale about the wise King Keykhosrow who owned a grand treasury filled with treasures from across the Persian Empire. However, to prevent theft and ensure the safety of his wealth, King Keykhosrow's vault was sealed with a magical lock that could only be opened by solving a riddle.

The riddle involves two sacred numbers $$$a$$$ and $$$b$$$. To unlock the vault, the challenger must determine the smallest key number $$$m$$$ that satisfies two conditions:

- $$$m$$$ must be greater than or equal to at least one of $$$a$$$ and $$$b$$$.
- The remainder when $$$m$$$ is divided by $$$a$$$ must be equal to the remainder when $$$m$$$ is divided by $$$b$$$.

Only by finding the smallest correct value of $$$m$$$ can one unlock the vault and access the legendary treasures!

Input: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$), the number of test cases.

Each test case consists of a single line containing two integers $$$a$$$ and $$$b$$$ ($$$1 \leq a, b \leq 1000$$$).

Output: For each test case, print the smallest integer $$$m$$$ that satisfies the conditions above.

Examples:
input:
2
4 6
472 896
output:
12
52864

Note: In the first test case, you can see that:

- $$$4 \bmod 4 = 0$$$ but $$$4 \bmod 6 = 4$$$
- $$$5 \bmod 4 = 1$$$ but $$$5 \bmod 6 = 5$$$
- $$$6 \bmod 4 = 2$$$ but $$$6 \bmod 6 = 0$$$
- $$$7 \bmod 4 = 3$$$ but $$$7 \bmod 6 = 1$$$
- $$$8 \bmod 4 = 0$$$ but $$$8 \bmod 6 = 2$$$
- $$$9 \bmod 4 = 1$$$ but $$$9 \bmod 6 = 3$$$
- $$$10 \bmod 4 = 2$$$ but $$$10 \bmod 6 = 4$$$
- $$$11 \bmod 4 = 3$$$ but $$$11 \bmod 6 = 5$$$

Buggy Code:
import java.util.Scanner;

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

        while (t-- > 0) {
            long a = sc.nextLong();
            long b = sc.nextLong();

            long gcd = gcd(a, b);
            long lcm = (a * b) / gcd;

            long minVal = Math.max(a, b);
            long m = minVal;

            while (m % a != m % b) {
                m += gcd;
            }

            System.out.println(m);
        }
    }

    // Helper method to calculate GCD using Euclidean algorithm
    public static long gcd(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
}

// kDMGiKNJqn7LsacEPAf

**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
1 1
1000 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: King Keykhosrow's Mystery
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a tale about the wise King Keykhosrow who owned a grand treasury filled with treasures from across the Persian Empire. However, to prevent theft and ensure the safety of his wealth, King Keykhosrow's vault was sealed with a magical lock that could only be opened by solving a riddle.

The riddle involves two sacred numbers $$$a$$$ and $$$b$$$. To unlock the vault, the challenger must determine the smallest key number $$$m$$$ that satisfies two conditions:

- $$$m$$$ must be greater than or equal to at least one of $$$a$$$ and $$$b$$$.
- The remainder when $$$m$$$ is divided by $$$a$$$ must be equal to the remainder when $$$m$$$ is divided by $$$b$$$.

Only by finding the smallest correct value of $$$m$$$ can one unlock the vault and access the legendary treasures!

Input: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$), the number of test cases.

Each test case consists of a single line containing two integers $$$a$$$ and $$$b$$$ ($$$1 \leq a, b \leq 1000$$$).

Output: For each test case, print the smallest integer $$$m$$$ that satisfies the conditions above.

Examples:
input:
2
4 6
472 896
output:
12
52864

Note: In the first test case, you can see that:

- $$$4 \bmod 4 = 0$$$ but $$$4 \bmod 6 = 4$$$
- $$$5 \bmod 4 = 1$$$ but $$$5 \bmod 6 = 5$$$
- $$$6 \bmod 4 = 2$$$ but $$$6 \bmod 6 = 0$$$
- $$$7 \bmod 4 = 3$$$ but $$$7 \bmod 6 = 1$$$
- $$$8 \bmod 4 = 0$$$ but $$$8 \bmod 6 = 2$$$
- $$$9 \bmod 4 = 1$$$ but $$$9 \bmod 6 = 3$$$
- $$$10 \bmod 4 = 2$$$ but $$$10 \bmod 6 = 4$$$
- $$$11 \bmod 4 = 3$$$ but $$$11 \bmod 6 = 5$$$

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

typedef long long int ll;

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<string> vs;

typedef vector<vector<int>> vvi;
typedef vector<vector<ll>> vvll;
#include <numeric>
#define yes cout<<"YES\n";
#define no cout<<"NO\n";
#define pb push_back
#define mp make_pair
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define loop(i,a,b) for(int i=(a);i<(b);i++)
const int mod = 1e9+7;

#ifndef ONLINE_JUDGE
    #define dbg(x) cout << #x << ":- " << x << "\n"
#else
    #define dbg(x)
#endif

ll gcd(ll a, ll b) {
    while (b != 0) {
        ll temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}
ll lcm(ll a, ll b) {
    return (a / gcd(a, b)) * b;
}
template <typename T> void ou(const T& value) { cout << value << " "; }
template <typename T> void ou(const vector<T>& vec) {  for (const auto& item : vec) print(item); cout<<endl; }
vll in(ll n) { vll vec(n); for (auto& item : vec) cin >> item; return vec; }
vll preSum(const vll& arr) { vll ps(arr.size()); for (int i = 0; i < arr.size(); ++i) ps[i] = (i == 0 ? arr[i] : ps[i-1] + arr[i]); return ps; }
vll sufSum(const vll& arr) { vll ss(arr.size()); for (int i = arr.size() - 1; i >= 0; --i) ss[i] = (i == arr.size() - 1 ? arr[i] : ss[i + 1] + arr[i]); return ss; }
vll sieve(ll n) { vll p; vector<bool> isP(n + 1, true); for (ll i = 2; i * i <= n; ++i) if (isP[i]) for (ll j = i * i; j <= n; j += i) isP[j] = false; for (ll i = 2; i <= n; ++i) if (isP[i]) p.push_back(i); return p; }
pair<ll, ll> cEO(vector<ll>& numbers) { ll even = 0, odd = 0; for (auto num : numbers) (num % 2 == 0 ? even : odd)++; return {even, odd}; }
ll p(ll a, ll b) { ll r = 1; for (; b; b >>= 1, a = (a % mod) * (a % mod) % mod) if (b & 1) r = (r % mod) * (a % mod) % mod; return r % mod; }
ll power(ll a, ll b, ll m) { ll result = 1; for (a %= m; b; b >>= 1, a = (a * a) % m) if (b & 1) result = (result * a) % m; return result; }
ll add(ll a, ll b, ll m) { return (a % m + b % m) % m; }
ll sub(ll a, ll b, ll m) { return (a % m - b % m + m) % m; }
ll mul(ll a, ll b, ll m) { return (a % m * b % m) % m; }
ll div(ll a, ll b, ll m) { return mul(a, power(b, m - 2, m), m); }

int main() {
    long long t;
    cin>>t;
    while(t--){
        ll a,b;
        cin>>a>>b;

        ll val=lcm(a,b);
        ll cnt=0;
     loop(i,min(a,b),val+1LL) {
          cnt=i;
       if((i%a)==(i%b)) break;

     }

       cout<<cnt<<endl;

              }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios where the code might fail. The key conditions to satisfy are:

1. \( m \) must be greater than or equal to at least one of \( a \) and \( b \).
2. The remainder when \( m \) is divided by \( a \) must be equal to the remainder when \( m \) is divided by \( b \).

The buggy code calculates the least common multiple (LCM) of \( a \) and \( b \) and then checks for the smallest \( m \) starting from \( \min(a, b) \) up to the LCM. However, this approach might fail in cases where the LCM is not the smallest \( m \) that satisfies the conditions.

Let's consider the following edge cases:
1. \( a \) and \( b \) are equal.
2. \( a \) and \( b \) are coprime (i.e., their greatest common divisor (GCD) is 1).
3. \( a \) is a multiple of \( b \) or vice versa.
4. \( a \) and \( b \) have a common divisor other than 1.

A particularly challenging test case is when \( a \) and \( b \) are coprime. In such cases, the LCM of \( a \) and \( b \) is \( a \times b \), which might be much larger than the smallest \( m \) that satisfies the conditions.

Here is a small-scale test input that includes a coprime pair and a pair where one number is a multiple of the other:

```plaintext
2
5 7
10 5
```

### Explanation:
1. **Test Case 1: \( a = 5 \) and \( b = 7 \)**
   - \( a \) and \( b \) are coprime.
   - The LCM of 5 and 7 is 35.
   - The smallest \( m \) that satisfies the conditions is 35, but the code might not find it efficiently.

2. **Test Case 2: \( a = 10 \) and \( b = 5 \)**
   - \( a \) is a multiple of \( b \).
   - The LCM of 10 and 5 is 10.
   - The smallest \( m \) that satisfies the conditions is 10.

This test input should help expose any inefficiencies or incorrect assumptions in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: King Keykhosrow's Mystery
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a tale about the wise King Keykhosrow who owned a grand treasury filled with treasures from across the Persian Empire. However, to prevent theft and ensure the safety of his wealth, King Keykhosrow's vault was sealed with a magical lock that could only be opened by solving a riddle.

The riddle involves two sacred numbers $$$a$$$ and $$$b$$$. To unlock the vault, the challenger must determine the smallest key number $$$m$$$ that satisfies two conditions:

- $$$m$$$ must be greater than or equal to at least one of $$$a$$$ and $$$b$$$.
- The remainder when $$$m$$$ is divided by $$$a$$$ must be equal to the remainder when $$$m$$$ is divided by $$$b$$$.

Only by finding the smallest correct value of $$$m$$$ can one unlock the vault and access the legendary treasures!

Input: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$), the number of test cases.

Each test case consists of a single line containing two integers $$$a$$$ and $$$b$$$ ($$$1 \leq a, b \leq 1000$$$).

Output: For each test case, print the smallest integer $$$m$$$ that satisfies the conditions above.

Examples:
input:
2
4 6
472 896
output:
12
52864

Note: In the first test case, you can see that:

- $$$4 \bmod 4 = 0$$$ but $$$4 \bmod 6 = 4$$$
- $$$5 \bmod 4 = 1$$$ but $$$5 \bmod 6 = 5$$$
- $$$6 \bmod 4 = 2$$$ but $$$6 \bmod 6 = 0$$$
- $$$7 \bmod 4 = 3$$$ but $$$7 \bmod 6 = 1$$$
- $$$8 \bmod 4 = 0$$$ but $$$8 \bmod 6 = 2$$$
- $$$9 \bmod 4 = 1$$$ but $$$9 \bmod 6 = 3$$$
- $$$10 \bmod 4 = 2$$$ but $$$10 \bmod 6 = 4$$$
- $$$11 \bmod 4 = 3$$$ but $$$11 \bmod 6 = 5$$$

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define le cout<<endl
#define yes cout<<"YES"
#define no cout<<"NO"
const int MOD = 1e9+7;

void solve(){
int a,b;
cin>>a>>b;
int low = min(a,b);
while(1){
    if(low%a==low%b){cout<<low;le;return;}
    low++;
}

}

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

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

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

Think step by step.

Response:
```plaintext
1
1 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: King Keykhosrow's Mystery
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a tale about the wise King Keykhosrow who owned a grand treasury filled with treasures from across the Persian Empire. However, to prevent theft and ensure the safety of his wealth, King Keykhosrow's vault was sealed with a magical lock that could only be opened by solving a riddle.

The riddle involves two sacred numbers $$$a$$$ and $$$b$$$. To unlock the vault, the challenger must determine the smallest key number $$$m$$$ that satisfies two conditions:

- $$$m$$$ must be greater than or equal to at least one of $$$a$$$ and $$$b$$$.
- The remainder when $$$m$$$ is divided by $$$a$$$ must be equal to the remainder when $$$m$$$ is divided by $$$b$$$.

Only by finding the smallest correct value of $$$m$$$ can one unlock the vault and access the legendary treasures!

Input: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$), the number of test cases.

Each test case consists of a single line containing two integers $$$a$$$ and $$$b$$$ ($$$1 \leq a, b \leq 1000$$$).

Output: For each test case, print the smallest integer $$$m$$$ that satisfies the conditions above.

Examples:
input:
2
4 6
472 896
output:
12
52864

Note: In the first test case, you can see that:

- $$$4 \bmod 4 = 0$$$ but $$$4 \bmod 6 = 4$$$
- $$$5 \bmod 4 = 1$$$ but $$$5 \bmod 6 = 5$$$
- $$$6 \bmod 4 = 2$$$ but $$$6 \bmod 6 = 0$$$
- $$$7 \bmod 4 = 3$$$ but $$$7 \bmod 6 = 1$$$
- $$$8 \bmod 4 = 0$$$ but $$$8 \bmod 6 = 2$$$
- $$$9 \bmod 4 = 1$$$ but $$$9 \bmod 6 = 3$$$
- $$$10 \bmod 4 = 2$$$ but $$$10 \bmod 6 = 4$$$
- $$$11 \bmod 4 = 3$$$ but $$$11 \bmod 6 = 5$$$

Buggy Code:
n=int(input())
def m(n,a,b):
    m=max(a,b)
    for i in range(m,a*b+1):
        if i%a==i%b:
            return (i)
def king(n):
    r=[]
    for _ in range(n):
        a,b=map(int,input().split())
        r.append(m(n,a,b))
    for i in r:
        print(i)
(king(n))

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

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

Think step by step.

Response:
```plaintext
3
1 1
1000 1000
999 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: King Keykhosrow's Mystery
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a tale about the wise King Keykhosrow who owned a grand treasury filled with treasures from across the Persian Empire. However, to prevent theft and ensure the safety of his wealth, King Keykhosrow's vault was sealed with a magical lock that could only be opened by solving a riddle.

The riddle involves two sacred numbers $$$a$$$ and $$$b$$$. To unlock the vault, the challenger must determine the smallest key number $$$m$$$ that satisfies two conditions:

- $$$m$$$ must be greater than or equal to at least one of $$$a$$$ and $$$b$$$.
- The remainder when $$$m$$$ is divided by $$$a$$$ must be equal to the remainder when $$$m$$$ is divided by $$$b$$$.

Only by finding the smallest correct value of $$$m$$$ can one unlock the vault and access the legendary treasures!

Input: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$), the number of test cases.

Each test case consists of a single line containing two integers $$$a$$$ and $$$b$$$ ($$$1 \leq a, b \leq 1000$$$).

Output: For each test case, print the smallest integer $$$m$$$ that satisfies the conditions above.

Examples:
input:
2
4 6
472 896
output:
12
52864

Note: In the first test case, you can see that:

- $$$4 \bmod 4 = 0$$$ but $$$4 \bmod 6 = 4$$$
- $$$5 \bmod 4 = 1$$$ but $$$5 \bmod 6 = 5$$$
- $$$6 \bmod 4 = 2$$$ but $$$6 \bmod 6 = 0$$$
- $$$7 \bmod 4 = 3$$$ but $$$7 \bmod 6 = 1$$$
- $$$8 \bmod 4 = 0$$$ but $$$8 \bmod 6 = 2$$$
- $$$9 \bmod 4 = 1$$$ but $$$9 \bmod 6 = 3$$$
- $$$10 \bmod 4 = 2$$$ but $$$10 \bmod 6 = 4$$$
- $$$11 \bmod 4 = 3$$$ but $$$11 \bmod 6 = 5$$$

Buggy Code:
def solve():
    a,b = sorted(map(int,input().split()))
    for i in range(a,1000000):
        if i%a == i%b:
            print(i)
            return

for __ in range(int(input())):
    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
1 1
2 2
1 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: King Keykhosrow's Mystery
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a tale about the wise King Keykhosrow who owned a grand treasury filled with treasures from across the Persian Empire. However, to prevent theft and ensure the safety of his wealth, King Keykhosrow's vault was sealed with a magical lock that could only be opened by solving a riddle.

The riddle involves two sacred numbers $$$a$$$ and $$$b$$$. To unlock the vault, the challenger must determine the smallest key number $$$m$$$ that satisfies two conditions:

- $$$m$$$ must be greater than or equal to at least one of $$$a$$$ and $$$b$$$.
- The remainder when $$$m$$$ is divided by $$$a$$$ must be equal to the remainder when $$$m$$$ is divided by $$$b$$$.

Only by finding the smallest correct value of $$$m$$$ can one unlock the vault and access the legendary treasures!

Input: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$), the number of test cases.

Each test case consists of a single line containing two integers $$$a$$$ and $$$b$$$ ($$$1 \leq a, b \leq 1000$$$).

Output: For each test case, print the smallest integer $$$m$$$ that satisfies the conditions above.

Examples:
input:
2
4 6
472 896
output:
12
52864

Note: In the first test case, you can see that:

- $$$4 \bmod 4 = 0$$$ but $$$4 \bmod 6 = 4$$$
- $$$5 \bmod 4 = 1$$$ but $$$5 \bmod 6 = 5$$$
- $$$6 \bmod 4 = 2$$$ but $$$6 \bmod 6 = 0$$$
- $$$7 \bmod 4 = 3$$$ but $$$7 \bmod 6 = 1$$$
- $$$8 \bmod 4 = 0$$$ but $$$8 \bmod 6 = 2$$$
- $$$9 \bmod 4 = 1$$$ but $$$9 \bmod 6 = 3$$$
- $$$10 \bmod 4 = 2$$$ but $$$10 \bmod 6 = 4$$$
- $$$11 \bmod 4 = 3$$$ but $$$11 \bmod 6 = 5$$$

Buggy Code:
import java.util.Scanner;

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

            if (a <= b) {
                for (long i = a; i <= 1000000001; i++) {
                    if (i % a == i % b) {
                        System.out.println(i);
                        // continue;
                        break;
                    }
                }
            } else {
                for (long i = b; i <= 1000000001; i++) {
                    if (i % a == i % b) {
                        System.out.println(i);
                        // continue;
                        break;
                    }
                }
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
1 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: King Keykhosrow's Mystery
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a tale about the wise King Keykhosrow who owned a grand treasury filled with treasures from across the Persian Empire. However, to prevent theft and ensure the safety of his wealth, King Keykhosrow's vault was sealed with a magical lock that could only be opened by solving a riddle.

The riddle involves two sacred numbers $$$a$$$ and $$$b$$$. To unlock the vault, the challenger must determine the smallest key number $$$m$$$ that satisfies two conditions:

- $$$m$$$ must be greater than or equal to at least one of $$$a$$$ and $$$b$$$.
- The remainder when $$$m$$$ is divided by $$$a$$$ must be equal to the remainder when $$$m$$$ is divided by $$$b$$$.

Only by finding the smallest correct value of $$$m$$$ can one unlock the vault and access the legendary treasures!

Input: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 100$$$), the number of test cases.

Each test case consists of a single line containing two integers $$$a$$$ and $$$b$$$ ($$$1 \leq a, b \leq 1000$$$).

Output: For each test case, print the smallest integer $$$m$$$ that satisfies the conditions above.

Examples:
input:
2
4 6
472 896
output:
12
52864

Note: In the first test case, you can see that:

- $$$4 \bmod 4 = 0$$$ but $$$4 \bmod 6 = 4$$$
- $$$5 \bmod 4 = 1$$$ but $$$5 \bmod 6 = 5$$$
- $$$6 \bmod 4 = 2$$$ but $$$6 \bmod 6 = 0$$$
- $$$7 \bmod 4 = 3$$$ but $$$7 \bmod 6 = 1$$$
- $$$8 \bmod 4 = 0$$$ but $$$8 \bmod 6 = 2$$$
- $$$9 \bmod 4 = 1$$$ but $$$9 \bmod 6 = 3$$$
- $$$10 \bmod 4 = 2$$$ but $$$10 \bmod 6 = 4$$$
- $$$11 \bmod 4 = 3$$$ but $$$11 \bmod 6 = 5$$$

Buggy Code:
import java.util.Scanner;

public class AKingKeykhosrowMystery {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            long a = sc.nextLong();
            long b = sc.nextLong();
            long min = Math.min(a, b);

            // if (a <= b) {
            for (long i = min; i <= 1000000001; i++) {
                if (i % a == i % b) {
                    System.out.println(i);
                    // continue;
                    break;
                }
            }
            // }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 1
2 3
1000 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

It is guaranteed that the sum of the values $n \cdot m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

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

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
//using namespace __gnu_pbds;
#define F first
#define S second
#define ll long long
#define ld long double
#define ull unsigned long long
#define pb push_back
#define ins insert
#define FAST ios_base::sync_with_stdio(NULL);cin.tie(0); cout.tie(0)
#define TIME cerr<<double(clock()-st)/(double)CLOCKS_PER_SEC
#define repp(l,r) for(ll i=l;i<r;i++)
#define rep(l,r) for(ll j=l;j<r;j++)
#define C continue
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define yes cout << "YES\n"
#define no cout << "NO\n"
#define mid ((l+r)>>1)
#define le (at<<1)
#define ri (at<<1|1)
#define vi vector<ll>
#define pi pair<ll,ll>
#define tup tuple<ll,ll,ll>
//template<class T>
//using TREE=tree<T,null_type,less_equal<T>,rb_tree_tag,
//tree_order_statistics_node_update>;
const ll N = 2e5+35;
const ll M = 1e8;
const ll oo = 1e18+18;
ll t , n , m , x , y , k;
ll base = 31 , inv;
int dr[8]={-1,1,0,0,-1,-1,1,1};
int dc[8]={0,0,-1,1,-1,1,-1,1};
ll add(ll a,ll b){
    a %= M;
    b %= M;
    return (a+b)%M;
}
ll sub(ll a , ll b) {
    a %= M;
    b %= M;
    b += M;
    return (a+b)%M;
}
ll mul(ll a , ll b) {
    a %= M;
    b %= M;
    return (a*b)%M;
}
bool prime(ll x) {
    if(x==2)return 1;
    if(x==1||x%2==0)return 0;
    for(ll i=3;i*i<=x;i+=2){
        if(x%i==0)return 0;
    }
    return 1;
}
struct My_hash {
    static uint64_t splitmix64(uint64_t x){
        x+=0x9e3779b97f4a7c15;
        x=(x^(x>>30))*0xbf58476d1ce4e5b9;
        x=(x^(x>>27))*0x94d049bb133111eb;
        return x^(x>>31) ;
    }
};
//void push_back(char c , ll&h) {
//    h = (h*base+c-'a'+1)%M;
//}
//void pop_front(char c , ll&h , ll l) {
//    h = (h-(pw[l-1]*(c-'a'+1))%M+M)%M;
//}
//ll pw(ll m , ll base) {
//   ull res=1;
//    while(m){
//       if(m&1)res*=base;
//       base*=base;
//        m>>=1;
//    }
//   return res;
//}
ll fast(ll base,ll m){
    ll res=1;
    while(m){
        if(m&1)res*=base;
        base*=base;
        m>>=1;
    }
    return res;
}
vector<int>c[N],v[N];
int main() {
    FAST;
    cin>>t;
    while(t--) {
        cin >> n >> m;
        unordered_map<ll,bool>sum;
        int cur=0;
        repp(0, n) {
            rep(0, m) {
                cin>>x;
                c[i].pb(x);
            }
        }
        repp(0, n) {
            rep(0, m) {
                cin >> x;
                v[i].pb(x);
            }
        }
        unordered_map<int,int>is;
        repp(0,n){
            rep(0,m){
                if(i!=j)C;
                is[c[i][j]]=1;
                ll cnt=0;
                for(int k=0;k<n;k++)cnt+=c[k][j];
                for(int k=0;k<m;k++)cnt+=c[i][k];
                sum[cnt]=1;
            }
        }
        int a = 0, b = 0;
        int f=0;
        repp(0,n){
            rep(0,m){
                if(is[v[i][j]]==0)C;
                ll cnt=0;
                for(int k=0;k<n;k++)cnt+=v[k][j];
                for(int k=0;k<m;k++)cnt+=v[i][k];
                f|=sum[cnt]==0;
            }
        }
        cout<<(f?"NO\n":"YES\n");
        rep(0,n)v[j].clear(),c[j].clear();
    }
}
//template<class T> using ordered_set =tree<T, null_type, less_equal<T>, rb_tree_tag,tree_order_statistics_node_update> ;
//ordered_set<int> st;
//order_of_key (k) : Number of items strictly smaller than k .   (number)
//find_by_order(k) : K-th element in a set (counting from zero).    (iterator)
//srand(chrono::steady_clock::now().time_since_epoch().count());
//freopen("convention2.in", "r", stdin);
//freopen("convention2.out", "w", stdout);
//int dr[8]={-1,1,0,0,-1,-1,1,1};
//int dc[8]={0,0,-1,1,-1,1,-1,1};
//bool ok(int i , int j) {
//    return(i>=0&&j>=0&&i<n&&j<n);
//}
//bool prime(ll x) {
//    if(x<=2) return x - 1;
//    if((x+1)&1) return 0;
//    for(int i = 3 ; i*i<=x ; i+=2) {
//        if(x % i == 0) return 0;
//    }
//    return 1;
//}
//bool check(string s) {
//    int sz = s.size();
//    rep(0 , sz) {
//        if(s[j] != s[sz - 1 - j]) return 0;
//    }
//    return 1;
//}
//queue<int>q;
//vector<int>v[N];
//vector<int>vis(N),in(N);
//void dfs(int i) {
//    vis[i] = 1;
//    for(auto j : v[i]) {
//        if(!vis[j]) dfs(j);
//    }
//}
//
//void sieve() {
//    vis[0]=vis[1]=1;
//    for(int i = 2 ; i*i<=N; i++) {
//        if(vis[i]) continue;
//        for(int j = i*i ; j <= N ; j+=i) {
//            vis[j] = 1;
//        }
//    }
//}
//int fun(int x , int b) {
//    return ((s[x]-'0')^b);
//}
//ll add(ll a , ll b) {
//    return (a % M + b % M) % M;
//}
//ll mul(ll a , ll b) {
//    a %= M;
//    b %= M;
//    return (a * b) % M;
//}
//ll fast(ll m , ll base = 2) {
//    ll res = 1;
//    while(m) {
//        if(m&1) res = mul(res , base);
//        base = mul(base , base);
//        m >>= 1;
//    }
//    return res;
//}
//void topolgical() {
//    while(q.size()) {
//        int x = q.front();
//        q.pop();
//        cout << x << '\n';
//        for(auto j : v[x]) {
//            if(--in[j] == 0) q.push(j);
//        }
//    }
//}
//    ios_base::sync_with_stdio(0),
//    cin.tie(0), cout.tie(0);
//freopen( "input.txt" , "r" , stdin ) ;
//freopen( "output.txt" , "w" , stdout ) ;

**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 edge cases and scenarios that can maximize the failure probability. The buggy code seems to have issues with how it handles the sums of rows and columns, and it might not correctly handle the permutations and transformations.

### Step-by-Step Reasoning:

1. **Understanding the Bug:**
   - The code uses a sum of rows and columns to determine if the transformation is possible.
   - It might fail if the sums are the same but the matrices cannot be transformed into each other.

2. **Edge Cases:**
   - **Single Element Matrix:** This is trivial and should always be "YES."
   - **Small Matrices:** Matrices with a small number of elements can be easier to analyze.
   - **Matrices with the Same Sums but Different Permutations:** This is a key edge case to test.

3. **Example Test Case:**
   - Let's consider a 2x2 matrix where the sums of rows and columns are the same, but the matrices cannot be transformed into each other.

### Test Input:

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

### Explanation:
- **Matrix A:**
  ```
  1 2
  3 4
  ```
- **Matrix B:**
  ```
  2 1
  4 3
  ```
- **Sums:**
  - Row sums for A: 3, 7
  - Column sums for A: 3, 7
  - Row sums for B: 3, 7
  - Column sums for B: 3, 7
- **Transformation:**
  - Despite having the same sums, matrix A cannot be transformed into matrix B by swapping rows or columns.

This test case should expose the bug in the code, as it will incorrectly output "YES" when it should output "NO."


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

It is guaranteed that the sum of the values $n \cdot m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

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

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

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

#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define endl "\n"

#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x<<" "; _print(x); cerr<< "\n";
#else
#define debug(x)
#endif

template<class T> void _print(T a) { cerr << a; }
template<class T, class V> void _print(pair<T,V> p){ _print('{'); _print(p.first); _print(','); _print(p.second); _print('}'); }
template<class T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) { _print(i); cerr << " "; } cerr << "]"; }
template<class T> void _print(vector<vector<T>> v) { cerr << "\n"; for (auto i : v) { cerr << "[ "; _print(i); cerr << "]\n"; } }
template<class T> void _print(set<T> s) { cerr << "[ "; for (T i : s) { _print(i); cerr << " "; } cerr << "]"; }
template<class T, class V> void _print(map<T, V> m) { cerr << "\n"; for (auto i : m) { _print(i.first); cerr << " "; _print(i.second); cerr << "\n"; } }

typedef long long ll;

/*------------------------------------------------------------------------------------------------------------------------------------------------------*/

/*------------------------------------------------------------------------------------------------------------------------------------------------------*/

int main() {

#ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
  freopen("error.txt", "w", stderr);
#endif

  fastio();

  int t;
  cin>>t;

  while(t--){

    ll n,m;
    cin>>n>>m;

    vector<vector<ll>> a(n, vector<ll> (m));
    for(ll i=0;i<n;i++){
      for(ll j=0;j<m;j++){
        cin>>a[i][j];
      }
    }

    vector<ll> r1,c1;

    for(ll i=0;i<n;i++){
      ll curr = 0;
      for(ll j=0;j<m;j++){
        curr+=a[i][j];
      }
      r1.push_back(curr);
    }

    for(ll j=0;j<m;j++){
      ll curr = 0;
      for(ll i=0;i<n;i++){
        curr+=a[i][j];
      }
      c1.push_back(curr);
    }

    vector<vector<ll>> b(n, vector<ll> (m));
    for(ll i=0;i<n;i++){
      for(ll j=0;j<m;j++){
        cin>>b[i][j];
      }
    }

    vector<ll> r2,c2;

    for(ll i=0;i<n;i++){
      ll curr = 0;
      for(ll j=0;j<m;j++){
        curr+=b[i][j];
      }
      r2.push_back(curr);
    }

    for(ll j=0;j<m;j++){
      ll curr = 0;
      for(ll i=0;i<n;i++){
        curr+=b[i][j];
      }
      c2.push_back(curr);
    }

    vector<ll> x,y;
    for(auto i:r1){
      for(auto j:c1){
        x.push_back(i+j);
      }
    }

    for(auto i:r2){
      for(auto j:c2){
        y.push_back(i+j);
      }
    }

    sort(x.begin(),x.end());
    sort(y.begin(),y.end());

    bool res = true;

    for(ll i=0;i<x.size();i++){
      if(x[i]!=y[i]){
        res = false;
        break;
      }
    }

    if(res){
      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
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
2 3
4 1
```

### Explanation:
1. **First Test Case**:
   - Matrix `a`:
     ```
     1 2
     3 4
     ```
   - Matrix `b`:
     ```
     4 3
     2 1
     ```
   - This test case should return "YES" because swapping the rows and then the columns will transform `a` into `b`.

2. **Second Test Case**:
   - Matrix `a`:
     ```
     1 2
     3 4
     ```
   - Matrix `b`:
     ```
     4 3
     2 1
     ```
   - This test case is identical to the first one and should also return "YES".

3. **Third Test Case**:
   - Matrix `a`:
     ```
     1 2
     3 4
     ```
   - Matrix `b`:
     ```
     2 3
     4 1
     ```
   - This test case should return "NO" because it is impossible to transform `a` into `b` using only row and column swaps. The elements in the second row of `b` do not match any possible row or column permutation of `a`.

This input is designed to expose the bug in the provided buggy code by focusing on the specific conditions where the sum of rows and columns might lead to incorrect conclusions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

It is guaranteed that the sum of the values $n \cdot m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

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

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
//         اللهم لا سهل الا ما جعلته سهلا و انت تجعل الصعب ان شئت سهل
#include <bits/stdc++.h>

using namespace std;

#define Time cerr << "Time Taken: " << (float)clock() / CLOCKS_PER_SEC << " Secs" << "\n";
#define tolower(s) transform(s.begin(), s.end(), s.begin(), ::tolower)
#define ll long long
#define ull unsigned long long
#define ld long double
#define sz(x) int(x.size())
#define all(v) v.begin(), v.end()
#define getline(s) getline(cin >> ws, s)
#define rall(vec) vec.rbegin(), vec.rend()
#define unq(vec) sort(all(vec)); vec.resize(unique(all(vec)) - vec.begin());
#define find_s(s, x) s.find(x) != string::npos // search for x in s
#define ceil(a, b) (((a) / (b)) + ((a) % (b) ? 1 : 0))
#define fi first
#define se second
#define fixed(n) cout << fixed << setprecision(n);
#define Pair pair < ll, ll >
#define pb push_back
#define pf push_front
#define eb emplace_back

template < typename T = int > istream& operator >> (istream &in, vector < T > &v) {
    for (auto &x : v) in >> x;
    return in;
}

template < typename T = int > ostream& operator << (ostream &out, const vector < T > &v) {
    for (const T &x: v) out << x << ' ';
    return out;
}

void solve() {
    int n, m;   cin >> n >> m;
    vector < vector < ll > > x(n, vector < ll >(m));
    vector < vector < ll > > y(n, vector < ll >(m));

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

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

    map < vector < ll >, bool > mp;

    for(int i = 0; i < m; i++) {
        vector < ll > tmp;
        for(int j = 0; j < n; j++) {
            tmp.pb(x[j][i]);
        }
        sort(all(tmp));
        mp[tmp] = true;
    }

    for(int i = 0; i < m; i++) {
        vector < ll > tmp;
        for(int j = 0; j < n; j++) {
            tmp.pb(y[j][i]);
        }
        sort(all(tmp));
        if(!mp[tmp]) {
            return void(cout << "NO\n");
        }
    }

    cout << "YES\n";
}

int main(){
	ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout);
    #endif

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

It is guaranteed that the sum of the values $n \cdot m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

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

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
t=int(input())
while t!=0:
    class segment:
        def __init__(self,n):
            self.n=n
            self.p=[i for i in range(n+1)]
        def pa(self,x):
            if self.p[x]==x:
                return x
            else:
                y=self.pa(self.p[x])
                self.p[x]=y
                return y
        def merge(self,a,b):
            x=self.pa(a)
            y=self.pa(b)
            self.p[x]=y
        def same(self,a,b):
            return self.pa(a)==self.pa(b)

    t-=1
    n,m=map(int,input().split())

    arr=[]
    for i in range(n):
        arr.append(list(map(int,input().split())))
    nums=[]
    for i in range(n):
        nums.append(list(map(int,input().split())))
    # print(arr)
    # print(nums)
    r=segment(n*m)
    c=segment(n*m)
    for i in range(n):
        for j in range(m-1):
            r.merge(arr[i][j],arr[i][j+1])
    for i in range(m):
        for j in range(n-1):
            c.merge(arr[j][i],arr[j+1][i])
    ans=True
    for i in range(n):
        for j in range(m-1):
            ans=ans and r.same(nums[i][j],nums[i][j+1])
    for i in range(m):
        for j in range(n-1):
            ans=ans and c.same(nums[j][i],nums[j+1][i])
    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
1 2
3 4
2 1
4 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

It is guaranteed that the sum of the values $n \cdot m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

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

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
from collections import defaultdict as dd
def fxn(a,b):
    rowa=dd(int)
    rowb=dd(int)
    cola=dd(int)
    colb=dd(int)
    for i in range(len(a)):
        asum=sum(a[i])
        rowa[asum]+=1
        bsum=sum(b[i])
        rowb[bsum]+=1

    for j in range(len((a)[0])):
        curra=0
        currb=0
        for k in range(len(a)):
            curra+=a[k][j]
            currb+=b[k][j]
        cola[curra]+=1
        colb[currb]+=1

    return rowa==rowb and cola==colb

for _ in range(int(input())):
    n,m=list(map(int,input().split()))
    a=[]
    for i in range(n):
        a.append(list(map(int,input().split())))
    b=[]
    for j in range(n):
        b.append(list(map(int,input().split())))
    if (fxn(a,b)):
        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
1 2 3
4 5 6
7 8 9
1 3 2
4 6 5
7 9 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

It is guaranteed that the sum of the values $n \cdot m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

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

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
def solve():
    n, m = list(map(int, input().split()))
    a = []
    b = []
    for i in range(n):
        a.append(list(map(int, input().split())))
    for i in range(n):
        b.append(list(map(int, input().split())))
    # print(m, n)
    # print(a)
    # print(b)
    set_col = []
    for j in range(m):
        set_ = set()
        for elem in [a[i][j] for i in range(n)]:
            set_.add(elem)
        set_col.append(set_)

    allocation = {}
    for j in range(m):
        for i, s in enumerate(set_col):
            if b[0][j] in s:
                if i in allocation:
                    return "NO"
                allocation[i] = j
    for k, v in allocation.items():
        set_a = [a[i][k] for i in range(n)]
        set_b = [b[i][v] for i in range(n)]
        if sorted(set_a) != sorted(set_b):
            return "NO"
    return "YES"

for _ in range(int(input())):
    res = solve()
    print(res)

**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 3
1 2 3
4 5 6
3 1 2
6 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

It is guaranteed that the sum of the values $n \cdot m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

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

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

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

public class Main {
    //private static final int MOD = 1_000_000_007;

    static InputReader scanner;
    static PrintWriter out;

    public static void main(String[] args) {
        final InputStream inputStream = System.in;
        final OutputStream outputStream = System.out;
        scanner = new InputReader(inputStream);
        out = new PrintWriter(outputStream);

        int t = scanner.nextInt();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }

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

    private static void solve() {
        final int n = scanner.nextInt();
        final int m = scanner.nextInt();

        final int[] minOnes = new int[n];
        final int[] minTwos = new int[n];

        final int[] rowMinTwosMap = new int[n*m+1];

        final int[][] arr = new int[n][m];
        final int[][] brr = new int[n][m];

        for(int i=0; i<n; i++){
            int min = Integer.MAX_VALUE;
            for(int j=0; j<m; j++){
                arr[i][j] = scanner.nextInt();
                min = Math.min(arr[i][j], min);
            }
            minOnes[i] = min;
        }

        for(int i=0; i<n; i++){
            int min = Integer.MAX_VALUE;
            for(int j=0; j<m; j++){
                brr[i][j] = scanner.nextInt();
                min = Math.min(brr[i][j], min);
            }
            minTwos[i] = min;
            rowMinTwosMap[min] = i;
        }

        Arrays.sort(minTwos);
        final int[] countArr = new int[n*m+1];
        final int[] mainMap = new int[m*n+1];
        final int[] indexMap = new int[m];

        // first mapping
        int val = Arrays.binarySearch(minTwos, minOnes[0]);
        if(val < 0){
            out.println("NO");
            return;
        }

        int row = rowMinTwosMap[minTwos[val]];

        for(int j=0; j<m; j++) {
            countArr[arr[0][j]]++;
            mainMap[arr[0][j]] = j;
        }

        for(int j=0; j<m; j++) {
            countArr[brr[row][j]]++;

            if( countArr[brr[row][j]] != 2) {
                out.println("NO");
                return;
            }

            indexMap[ mainMap[brr[0][j]] ] = j;
        }

        for(int rOne = 1; rOne<n; rOne++){
            int rTwoIndex = Arrays.binarySearch(minTwos, minOnes[rOne]);
            if(rTwoIndex < 0){
                out.println("NO");
                return;
            }

            int rTwo = rowMinTwosMap[minTwos[rTwoIndex]];

            for(int j=0; j<m; j++) {
                countArr[arr[rOne][j]]++;
                mainMap[arr[rOne][j]] = j;
            }

            for(int j=0; j<m; j++) {
                countArr[brr[rTwo][j]]++;

                if( countArr[brr[rTwo][j]] != 2) {
                    out.println("NO");
                    return;
                }

                if( indexMap[mainMap[brr[rOne][j]]] != j){
                    out.println("NO");
                    return;
                }

            }
        }
        out.println("YES");
    }

    static class Pair{
        public int first;
        public int second;
        private final int hashCode;

        public Pair(int first, int second) {
            this.first = first;
            this.second = second;
            this.hashCode = Objects.hash(first,second);
        }

        @Override
        public int hashCode() {
            return hashCode;
        }

        @Override
        public boolean equals(Object obj) {
            if(this == obj) return true;

            if(obj == null || obj.getClass() != this.getClass()) return false;

            Pair pair = (Pair) obj;
            return pair.first == first && pair.second == this.second;
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
        public 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 provided code, we need to craft a test case that highlights the issue. The code appears to be checking the minimum values in each row and using these to map the rows of the two matrices. However, this approach might fail in cases where the minimum values do not uniquely identify the rows or where the order of elements in the rows is crucial.

### Step-by-Step Analysis:
1. **Matrix Permutation Properties**: Both matrices are permutations of integers from 1 to \( n \cdot m \). This means each number from 1 to \( n \cdot m \) appears exactly once in each matrix.
2. **Row and Column Swaps**: The operations allowed are row and column swaps. These operations do not change the set of elements in the matrix, only their positions.
3. **Minimum Values in Rows**: The code uses the minimum values in each row to map the rows of the two matrices. This can be problematic if multiple rows have the same minimum value.

### Constructing a Test Case:
- **Equal Minimum Values in Rows**: Create a matrix where multiple rows have the same minimum value. This can cause the code to incorrectly map rows.
- **Different Order of Elements**: Ensure that the elements in the rows are in a different order in the target matrix to make it non-trivial to map.

### Example Test Case:
```plaintext
1
3 3
1 2 3
4 5 6
7 8 9
1 4 7
2 5 8
3 6 9
```

### Explanation:
- **Matrix A**:
  ```
  1 2 3
  4 5 6
  7 8 9
  ```
- **Matrix B**:
  ```
  1 4 7
  2 5 8
  3 6 9
  ```

In this test case:
- Both matrices are permutations of the numbers from 1 to 9.
- The minimum values in the rows of Matrix A are 1, 4, and 7, which are the same as the minimum values in the rows of Matrix B.
- However, the order of elements in Matrix B is different from Matrix A, requiring column swaps to achieve the transformation.

### Expected Output:
- **Output**: "YES"

This test case should expose the bug if the code incorrectly maps the rows based on the minimum values and fails to consider the order of elements.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

It is guaranteed that the sum of the values $n \cdot m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

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

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class PermutationRowColoumn {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int[][]arr1 = new int[n][m];
            int[][]arr2 = new int[n][m];
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < m; k++) {
                    arr1[j][k] = sc.nextInt();
                }
            }
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < m; k++) {
                    arr2[j][k] = sc.nextInt();
                }
            }
            String ans = helper(arr1,arr2);
            System.out.println(ans);
        }
    }
    public static String helper(int[][]arr1,int[][]arr2){
        ArrayList<Integer>list1 = new ArrayList<>();//for rows sum arr1
        ArrayList<Integer>list2 = new ArrayList<>();// for column sum arr1
        ArrayList<Integer>list3 = new ArrayList<>(); // for rows sum arr2
        ArrayList<Integer>list4 = new ArrayList<>(); // for column sum arr2
        ArrayList<ArrayList<Integer>>ans1 =new ArrayList<>();
        ArrayList<ArrayList<Integer>>ans2 =new ArrayList<>();
        ArrayList<ArrayList<Integer>>ans3 =new ArrayList<>();
        ArrayList<ArrayList<Integer>>ans4 =new ArrayList<>();
        for (int i = 0; i < arr1.length; i++) {
            int sum =0;
            for (int j = 0; j < arr1[0].length; j++) {
                sum = sum + arr1[i][j];
            }
            list1.add(sum);//having sum of all rows values.
        }
        for (int i = 0; i < arr1[0].length; i++) {
            int sum =0;
            for (int j = 0; j < arr1.length; j++) {
                sum = sum + arr1[j][i];
            }
            list2.add(sum); // having sum of all column values.
        }
        for (int i = 0; i < arr2.length; i++) {
            int sum =0;
            for (int j = 0; j < arr2[0].length; j++) {
                sum = sum + arr2[i][j];
            }
            list3.add(sum);
        }
        for (int i = 0; i < arr2[0].length; i++) {
            int sum =0;
            for (int j = 0; j < arr2.length; j++) {
                sum = sum + arr2[j][i];
            }
            list4.add(sum);
        }
        Collections.sort(list1);
        Collections.sort(list2);
        Collections.sort(list3);
        Collections.sort(list4);
        ans1.add(list1);
        ans2.add(list2);
        ans3.add(list3);
        ans4.add(list4);
        if(ans1.equals(ans3) && ans2.equals(ans4)){
            return "YES";
        }else{
            return "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
2
2 2
1 2
3 4
1 2
3 4
2 2
1 2
3 4
2 1
4 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

It is guaranteed that the sum of the values $n \cdot m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

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

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

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

void solve(){
    int n, m; cin>>n>>m;

    int a[n][m], b[n][m];
    map<int, int>row, col;
    for(int i=0; i<n; i++){
        for(int j=0; j<m; j++){
            cin>>a[i][j];
            if(i == 0) col[a[i][j]] = j;
            if(j == 0) row[a[i][j]] = i;
        }
    }

    // for(auto it : row) cout<<it.first<<" r "<<it.second<<endl;
    // for(auto it : col) cout<<it.first<<" c "<<it.second<<endl;

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

    vector<unordered_set<int>>st(n);
    for(int i=0; i<n; i++){
        unordered_set<int>tmp;
        int ind = -1;
        for(int j=0; j<m; j++){
            if(row.count(b[i][j])) ind = row[b[i][j]];
            tmp.insert(a[i][j]);
        }
        // cout<<i<<" first "<<ind<<endl;
        st[ind] = tmp;
    }

    bool flag = true;

    for(int i=0; i < n && m > 1; i++){
        for(int j=0; j<m; j++){
            if(st[i].find(b[i][j]) == st[i].end()){
                // cout<<i<<" rc "<<j<<endl;
                flag = false;
                break;
            }
        }
        if(!flag) break;
    }

    if(!flag){
        cout<<"NO"<<endl;
        return;
    }

    st.clear();
    st.resize(m);
    for(int i=0; i<m; i++){
        unordered_set<int>tmp;
        int ind = -1;
        for(int j=0; j<n; j++){
            if(col.count(b[j][i])) ind = col[b[j][i]];
            tmp.insert(a[j][i]);
        }
        st[ind] = tmp;
    }

    for(int j=0; j < m && n > 1; j++){
        for(int i=0; i<n; i++){
            if(st[j].find(b[i][j]) == st[j].end()){
                // cout<<i<<" cc "<<j<<endl;
                flag = false;
                break;
            }
        }
        if(!flag) break;
    }

    if(flag) cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
}

int main() {
    int t;
    cin>>t;
    while(t--) solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

It is guaranteed that the sum of the values $n \cdot m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

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

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

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

#define inf numeric_limits<int>::max();
#define rinf numeric_limits<int>::min();
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define ll long long
#define rep(i, s, e) for(ll i=s; i<e; i++)
#define rrep(i, s, e) for(ll i=e-1; i>=s; i--)
#define pb push_back
#define endl "\n";

void solve() {
    int n, m; cin >> n >> m;

    multiset<int> s;

    vector<vector<int>> v1(n, vector<int>(m));
    vector<vector<int>> v2(n, vector<int>(m));

    ll sm, in;
    rep(i, 0, n) {
        sm = 0;
        rep(j, 0, m) {
            cin >> in;
            v1[i][j] = in;
            sm += v1[i][j];
        }
        s.insert(sm);
    }

    rep(i, 0, n) {
        sm = 0;
        rep(j, 0, m) {
            cin >> in;
            v2[i][j] = in;
            sm += v2[i][j];
        }
        auto pos = s.find(sm);
        if (pos != s.end()) s.erase(pos);
        else {
            cout << "NO" << endl;
            return ;
        }
    }

    s.clear();

    rep(i, 0, m) {
        sm = 0;
        rep(j, 0, n) {
            sm += v1[j][i];
        }
        s.insert(sm);
    }

    rep(i, 0, m) {
        sm = 0;
        rep(j, 0, n) {
            sm += v2[j][i];
        }
        auto pos = s.find(sm);
        if (pos != s.end()) s.erase(pos);
        else {
            // cout << sm << endl;
            cout << "NO" << endl;
            return ;
        }
    }
    cout << "YES" << endl;

}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    //cout << std::setprecision(12);

    // freopen("f.in", "r", stdin);
	// freopen("f.out", "w", stdout);

    ll t = 1;
    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
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
2 2
1 3
2 4
4 1
3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation of Rows and Columns
time_limit_ms: 3000
memory_limit_mb: 256
Description: You have been given a matrix $a$ of size $n$ by $m$, containing a permutation of integers from $1$ to $n \cdot m$.

A permutation of $n$ integers is an array containing all numbers from $1$ to $n$ exactly once. For example, the arrays $[1]$, $[2, 1, 3]$, $[5, 4, 3, 2, 1]$ are permutations, while the arrays $[1, 1]$, $[100]$, $[1, 2, 4, 5]$ are not.

A matrix contains a permutation if, when all its elements are written out, the resulting array is a permutation. Matrices $[[1, 2], [3, 4]]$, $[[1]]$, $[[1, 5, 3], [2, 6, 4]]$ contain permutations, while matrices $[[2]]$, $[[1, 1], [2, 2]]$, $[[1, 2], [100, 200]]$ do not.

You can perform one of the following two actions in one operation:

  * choose columns $c$ and $d$ ($1 \le c, d \le m$, $c \ne d$) and swap these columns;    * choose rows $c$ and $d$ ($1 \le c, d \le n$, $c \ne d$) and swap these rows.

You can perform any number of operations.

You are given the original matrix $a$ and the matrix $b$. Your task is to determine whether it is possible to transform matrix $a$ into matrix $b$ using the given operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The descriptions of the test cases follow.

The first line of each test case description contains $2$ integers $n$ and $m$ ($1 \le n, m \le n \cdot m \le 2 \cdot 10^5$) — the sizes of the matrix.

The next $n$ lines contain $m$ integers $a_{ij}$ each ($1 \le a_{ij} \le n \cdot m$). It is guaranteed that matrix $a$ is a permutation.

The next $n$ lines contain $m$ integers $b_{ij}$ each ($1 \le b_{ij} \le n \cdot m$). It is guaranteed that matrix $b$ is a permutation.

It is guaranteed that the sum of the values $n \cdot m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output "YES" if the second matrix can be obtained from the first, and "NO" otherwise.

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

Examples:
input:
7
1 1
1
1
2 2
1 2
3 4
4 3
2 1
2 2
1 2
3 4
4 3
1 2
3 4
1 5 9 6
12 10 4 8
7 11 3 2
1 5 9 6
12 10 4 8
7 11 3 2
3 3
1 5 9
6 4 2
3 8 7
9 5 1
2 4 6
7 8 3
2 3
1 2 6
5 4 3
6 1 2
3 4 5
1 5
5 1 2 3 4
4 2 5 1 3
output:
YES
YES
NO
YES
YES
NO
YES

Note: In the second example, the original matrix looks like this:

$ \begin{pmatrix} 1 & 2 \\\ 3 & 4 \end{pmatrix} $

By swapping rows $1$ and $2$, it becomes:

$ \begin{pmatrix} 3 & 4 \\\ 1 & 2 \end{pmatrix} $

By swapping columns $1$ and $2$, it becomes equal to matrix $b$:

$ \begin{pmatrix} 4 & 3 \\\ 2 & 1 \end{pmatrix} $

Buggy Code:
t=int(input())
while t!=0:
    class dsu:
        def __init__(self,n):
            self.n=n
            self.p=[i for i in range(n+1)]
            self.rank=[0]*(n+1)
        def pa(self,x):
            if self.p[x]==x:
                return x
            else:
                y=self.pa(self.p[x])
                self.p[x]=y
                return y
        def merge(self,a,b):
            x=self.pa(a)
            y=self.pa(b)
            if x!=y:
                if self.rank[x]>self.rank[y]:
                    self.p[y]=x
                elif self.rank[x]<self.rank[y]:
                    self.p[x]=y
                else:
                    self.p[y]=x
                    self.rank[x]+=1

        def same(self,a,b):
            return self.pa(a)==self.pa(b)

    t-=1
    n,m=map(int,input().split())

    arr=[]
    for i in range(n):
        arr.append(list(map(int,input().split())))
    nums=[]
    for i in range(n):
        nums.append(list(map(int,input().split())))
    # print(arr)
    # print(nums)
    r=dsu(n*m)
    c=dsu(n*m)
    for i in range(n):
        for j in range(m-1):
            r.merge(arr[i][j],arr[i][j+1])
    for i in range(m):
        for j in range(n-1):
            c.merge(arr[j][i],arr[j+1][i])
    ans=True
    for i in range(n):
        for j in range(m-1):
            ans=ans and r.same(nums[i][j],nums[i][j+1])
    for i in range(m):
        for j in range(n-1):
            ans=ans and c.same(nums[j][i],nums[j+1][i])
    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:
To expose the bug in the given code, we need to focus on edge cases that might cause the algorithm to fail. The code uses Disjoint Set Union (DSU) to track the connected components of elements in the matrix. The bug might be related to how the DSU is initialized or how the merging and checking of components is performed.

### Step-by-Step Analysis:
1. **Initialization of DSU:**
   - The DSU is initialized with `n * m` elements, which is correct for the size of the matrix.
   - However, the elements in the matrix are from 1 to `n * m`, not 0 to `n * m - 1`. This could lead to issues if the DSU is not properly handling these values.

2. **Merging and Checking:**
   - The merging and checking operations are performed based on the elements in the matrix.
   - The code assumes that the elements in the matrix are 0-indexed, which is incorrect. This could lead to out-of-bounds errors or incorrect merging.

### Test Input:
To expose this bug, we can create a test case where the elements in the matrix are from 1 to `n * m`, and the matrices are such that the DSU should correctly identify the connected components.

### Test Case:
```plaintext
1
2 2
1 2
3 4
2 1
4 3
```

### Explanation:
- **Matrix `a`:**
  - `1 2`
  - `3 4`
- **Matrix `b`:**
  - `2 1`
  - `4 3`

### Expected Output:
- **Output: "YES"**
- **Explanation:**
  - By swapping the columns, we can transform matrix `a` into matrix `b`.
  - The DSU should correctly identify that the elements in the same row or column are connected.

### Why This Test Case Exposes the Bug:
- The elements in the matrices are from 1 to 4, which is the correct range for a 2x2 matrix.
- If the DSU is not properly handling 1-based indices, it will fail to correctly merge or check the connected components, leading to an incorrect output.

This test case should help identify if the DSU is correctly handling the 1-based indices of the matrix elements.