**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
// ConsoleApplication2.cpp : This file contains the 'main' function. Program execution begins and ends there.
//
#include "bits/stdc++.h"

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define repll(i, n) for (long long i = 0; i < (n); ++i)
#define rep1(i, n) for (int i = 1; i < (n); ++i)
#define rep1n(i, n) for (int i = 1; i <= (n); ++i)
#define repr(i, n) for (int i = (n) - 1; i >= 0; --i)
//#define pb push_back
//#define eb emplace_back
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define each(x, a) for (auto &x : a)
#define ar array
#define vec vector
#define range(i, n) rep(i, n)

#define debug1(var) if (verbose) cout << "var=" << var << "\n";

#define dbg(label,...) if (verbose) cerr << label << " (" << #__VA_ARGS__ << "):", debug_out(__VA_ARGS__)

using namespace std;

void debug_out() { cout << endl; }
template <typename head, typename... tail> void debug_out(head h, tail... t) {
  cout << h << ' ';
  debug_out(t...);
}

using ll = long long;
using ull = unsigned long long;
using ld = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;

using vi = vector<int>;
using vl = vector<ll>;
using vpi = vector<pair<int, int>>;
using vvi = vector<vi>;

int Bit(int mask, int b) { return (mask >> b) & 1; }

int globalt=0;

// Modular functions
int mod = 1e9 + 7;

int add(int a, int b) {
    return a + b >= mod ? a + b - mod : a + b;
}
int sub(int a, int b) {
    return a >= b ? a - b : a + mod - b;
}
int mul(int a, int b) {
    return 1ll * a * b % mod;
}
int inv(int x) {
    return x == 1 ? 1 : sub(0, mul(mod / x, inv(mod % x)));
}

int BitCount(unsigned int u)
{
         unsigned int uCount;

         uCount = u - ((u >> 1) & 033333333333) - ((u >> 2) & 011111111111);
         return ((uCount + (uCount >> 3)) & 030707070707) % 63;
}

int myBits(int num) {
    int res = 0;
    while (num > 0) {
        num >>= 1;
        res++;
    }
    return res;
}

int myBitsLL(ll num) {
    int res = 0;
    while (num > 0) {
        num >>= 1;
        res++;
    }
    return res;
}

const ll INF = 2e18;
const int INFi = 1e9;

void printVector(vector<ll> arr)
{
    for (int i: arr)
        cout << i << "##";
    cout << endl;
}

int secDegEqIntPos(ll a, ll b, ll c, double* sol1, double* sol2)
{
    int res=0;
    double disc=1.0*b*b-4*a*c;
    //cout << "a=" << a << " b=" << b << " c=" << c  << "\n";
    //cout << "disc=" << disc << "\n";
    if (disc<0) return(0);
    if (disc==0) {
            *sol1=-1.0*b/(2*a);
            *sol2=0.0;
            return(1);
    }
    *sol1=(-1.0*b+sqrt(disc))/(2.0*a);
    *sol2=(-1.0*b-sqrt(disc))/(2.0*a);

    //cout << "intermediate " << (*sol1) << " " << (*sol2)  << "\n";

    // check if integer solution exists
    ll a1=ceil(*sol1);
    ll a1sol=a*(a1*a1)+b*a1+c;
    if (a1sol==0) *sol1=a1;
    else {
        ll a2=floor(*sol1);
        ll a1sol=a*(a2*a2)+b*a2+c;
        if (a1sol==0) *sol1=a2;
    }
    a1=ceil(*sol2);
    a1sol=a*(a1*a1)+b*a1+c;
    if (a1sol==0) *sol2=a1;
    else {
        ll a2=floor(*sol2);
        ll a1sol=a*(a2*a2)+b*a2+c;
        if (a1sol==0) *sol2=a2;
    }

    return(2);

}

typedef pair<int,int> mypoint;   // points for indexed list
// function to use for sorting from min to max
bool sortboolMinToMax(mypoint a1, mypoint a2)
{
    if (a1.first!=a2.first)
        return(a1.first<a2.first);
    else return(a1.second<a2.second);
}

// function to use for sorting from max to min
bool sortboolMaxToMin(mypoint a1, mypoint a2)
{
    if (a1.first!=a2.first)
        return(a1.first>a2.first);
    else return(a1.second>a2.second);
}

typedef struct {int first; int second; int firstC; int secondC;} mypair;

void solve() {

    int verbose=0;

    int n,m;
    cin >> n >> m;
    vector<vector<int>> a;
    vector<vector<int>> b;
    int vCol[m];
    int vRow[n];
    int auxi=0;

    rep(i,n) {
        vector<int> aux;
        rep(j,m) {
            cin >> auxi;
            aux.push_back(auxi);
            //dbg("P1",auxi);
        }
        a.push_back(aux);
    }
    rep(i,n) {
        vector<int> aux;
        rep(j,m) {
            cin >> auxi;
            aux.push_back(auxi);
            //dbg("P2",auxi);
        }
        b.push_back(aux);
    }

    // identify columns to change
    set<int> colTC;
    set<int> rowTC;

    int a1=0;
    int b1=0;
    int c1=0;

    int turn=0;
    int counter=0;
    int mFinal=0;

    while(counter<10) {
        colTC.clear();
        rowTC.clear();

        rep(i,n) vRow[i]=0;
        rep(i,m) vCol[i]=0;

        mFinal=1;

        rep(i,n) {
            rep(j,m) {
                a1=a[i][j] ^ b[i][j];
                b1=a1 & b[i][j];
                c1=a1 & (~b[i][j]);

                if (b1) {
                    mFinal=0;
                    colTC.insert(j);
                    vCol[j]=vCol[j]|b1;
                }
                if (c1) {
                    mFinal=0;
                    rowTC.insert(i);
                    vRow[i]=vRow[j]|c1;
                }

                dbg("I1",i,j,a[i][j],b[i][j],a1,b1,c1);
            }
        }

        if (verbose) {
            cout << "rows to change\n";
            for (int val:rowTC) cout << val << " ";
            cout << "\n";
            cout << "columns to change\n";
            for (int val:colTC) cout << val << " ";
            cout << "\n";
        }

        if (mFinal) break;

        if (turn==0) {
            // apply columns
            for (int cindex:colTC) {
                rep(i,n) a[i][cindex]=a[i][cindex]|vCol[cindex];
            }
            turn=1;
        } else {
            // apply rows
            for (int rindex:rowTC) {
                rep(i,m) a[rindex][i]=a[rindex][i]&(~vRow[rindex]);
            }
            turn=0;
        }
        counter++;
    }
    //cout << "COUNTER=" << counter << "\n";
    if (counter<10) cout << "Yes\n";
    else cout << "No\n";
}

signed main() {

    // freopen("C:\\cpp\\inputFiles\\input.txt","r",stdin);

    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(12) << fixed;
    int t = 1;
    cin >> t;
    if (t==0) {
            freopen("C:\\cpp\\inputFiles\\input.txt","r",stdin);
            cin >> t;
    }

    rep(i, t) {
        globalt=i+1;
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, let's consider the following steps:

1. **Understand the Bug**: The code seems to have a logical error in how it processes the bitwise operations to transform matrix A into matrix B. Specifically, the code might not correctly handle cases where the bitwise operations need to be applied in a specific order or where the values of `x` need to be carefully chosen.

2. **Identify Edge Cases**: Edge cases often include:
   - Matrices with very small dimensions (1x1, 1x2, 2x1, 2x2).
   - Matrices with all zeros or all ones.
   - Matrices where the transformation is possible but requires a specific sequence of operations.
   - Matrices where the transformation is not possible.

3. **Construct a Test Case**: Let's construct a test case where the transformation is possible but requires a specific sequence of operations. This will help in exposing the logical error in the code.

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

### Explanation:
- **Matrix A**:
  ```
  1 2
  3 4
  ```
- **Matrix B**:
  ```
  3 3
  1 2
  ```

### Steps to Transform:
1. **Column Operation**:
   - Apply OR operation on column 1 with `x = 3`:
     ```
     1 2
     3 4
     =>
     3 2
     3 4
     ```
2. **Row Operation**:
   - Apply AND operation on row 1 with `x = 3`:
     ```
     3 2
     3 4
     =>
     3 2
     3 4
     ```
3. **Row Operation**:
   - Apply AND operation on row 2 with `x = 1`:
     ```
     3 2
     3 4
     =>
     3 2
     1 0
     ```
4. **Column Operation**:
   - Apply OR operation on column 2 with `x = 3`:
     ```
     3 2
     1 0
     =>
     3 3
     1 2
     ```

This test case requires a specific sequence of operations to transform matrix A into matrix B. If the code does not handle this sequence correctly, it will fail to transform the matrices and output "No" instead of "Yes".

### Output:
```plaintext
1
2 2
1 2
3 4
3 3
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define int long long
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
#define pnl cout << "\n"
#ifndef ONLINE_JUDGE
#define dbg(x) cerr << (#x) << " " << (x) << "\n"
#else
#define dbg(x) 42
#endif

using namespace std;
using namespace __gnu_pbds;

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
using vi = vector<int>;
using vvi = vector<vi>;
using ii = array<int, 2>;
using vii = vector<ii>;

template <typename T>
istream &operator>>(istream &in, vector<T> &a) {
    for (auto &x : a) in >> x;
    return in;
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> &a) {
    for (auto &x : a) out << x << " ";
    return out;
}

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

void solve(int testcase) {
    int n, m; cin >> n >> m;
    vvi a(n, vi(m)), b(n, vi(m)); cin >> a >> b;

    vector<vector<bitset<32>>> A(n, vector<bitset<32>>(m)), B(n, vector<bitset<32>>(m));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            A[i][j] = a[i][j];
            B[i][j] = b[i][j];
        }
    }

    // (n + m) nodes
    // directed edges from a->b denotes that a must happen before b
    // now there should a topo sort such that all edges are satisfied
    // no cycle should be there (sufficient condition)

    vector<set<int>> adj(n + m);

    // OR with 1 makes it 1
    // AND with 0 makes it 0
    // Otherwise, nothing changes

    auto is_cycle = [&]() {
        vi vis(n + m); // 0: not visited, 1: visited, 2: processed
        function<bool(int)> dfs = [&](int u) {
            vis[u] = 1;
            for (int v : adj[u]) {
                if (vis[v] == 1) {
                    return true;
                }
                if (vis[v] == 0 && dfs(v)) {
                    return true;
                }
            }
            vis[u] = 2;
            return false;
        };
        for (int i = 0; i < n + m; i++) {
            if (vis[i] == 0 && dfs(i)) {
                return true;
            }
        }
        return false;
    };

    for (int b = 0; b < 32; b++) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (A[i][j][b] == B[i][j][b]) {
                    // ok need to handle here
                    continue;
                }
                if (A[i][j][b] == 1) {
                    // 1 to 0
                    // or then and
                    adj[j + n].insert(i);
                } else {
                    // 0 to 1
                    // and then or
                    adj[i].insert(j + n);
                }
            }
        }
        if (is_cycle()) {
            cout << "No\n";
            return;
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (A[i][j][b] == B[i][j][b]) {
                    // ok need to handle here
                    if (A[i][j][b] == 1) {
                        // 1 to 1
                        // there should not be -> or then and
                        if (adj[j + n].find(i) != adj[j + n].end()) {
                            cout << "No\n";
                            return;
                        }
                    } else {
                        // 0 to 0
                        // there should not be -> or then and
                        if (adj[i].find(j + n) != adj[i].end()) {
                            cout << "No\n";
                            return;
                        }
                    }
                    continue;
                }
            }
        }
        adj = vector<set<int>>(n + m);
    }

    cout << "Yes\n";
}

int32_t main() {
    using namespace chrono;
    ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
    auto start = high_resolution_clock::now();
    int T = 1; cin >> T;
    for (int t = 1; t <= T; t++) {
        solve(t);
    }
    auto end = high_resolution_clock::now();
    auto duration = duration_cast<milliseconds>(end - start);
#ifndef ONLINE_JUDGE
    cerr << "Time: " << duration.count() << " ms\n";
#endif
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that are likely to cause the algorithm to fail. Here are the steps to consider:

1. **Edge Cases**:
   - Matrices with very small dimensions (e.g., 1x1, 1x2, 2x1).
   - Matrices with all elements being 0 or all elements being a large value like \(10^9\).
   - Matrices where the transformation is impossible due to the nature of bitwise operations.

2. **Specific Scenarios**:
   - Matrices where the transformation is possible but requires a specific sequence of operations.
   - Matrices where the transformation is impossible due to the constraints of bitwise operations.

3. **Test Case**:
   - Let's create a test case where the transformation is impossible. This will help us verify if the code correctly identifies such cases.

### Chosen Test Case
- **Dimensions**: 2x2
- **Matrix A**:
  - Row 1: 1 2
  - Row 2: 3 4
- **Matrix B**:
  - Row 1: 0 0
  - Row 2: 0 0

### Explanation
- **Matrix A**:
  - Row 1: 1 (binary: 01) and 2 (binary: 10)
  - Row 2: 3 (binary: 11) and 4 (binary: 100)
- **Matrix B**:
  - Row 1: 0 (binary: 00) and 0 (binary: 00)
  - Row 2: 0 (binary: 00) and 0 (binary: 00)

### Why This Test Case?
- **Bitwise AND and OR Operations**:
  - To transform the elements of Matrix A to all zeros, we would need to AND all elements with 0. However, the AND operation can only set bits to 0, not to 1.
  - The OR operation can only set bits to 1, not to 0.
- **Impossibility**:
  - It is impossible to transform any non-zero element to 0 using only AND and OR operations without affecting other elements in the same row or column.

### Test Input
```plaintext
1
2 2
1 2
3 4
0 0
0 0
```

This test case should expose the bug in the provided code by correctly identifying that the transformation is impossible.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
import sys
import threading

def main():
    import sys

    def input_gen():
        for token in sys.stdin.read().split():
            yield token

    tokens = input_gen()
    try:
        t = int(next(tokens))
    except StopIteration:
        t = 0

    for _ in range(t):
        try:
            n = int(next(tokens))
            m = int(next(tokens))
        except StopIteration:
            n = m = 0

        A = []
        for _ in range(n):
            row = []
            for _ in range(m):
                try:
                    row.append(int(next(tokens)))
                except StopIteration:
                    row.append(0)
            A.append(row)

        B = []
        for _ in range(n):
            row = []
            for _ in range(m):
                try:
                    row.append(int(next(tokens)))
                except StopIteration:
                    row.append(0)
            B.append(row)

        C = [0] * m  # Column OR masks
        for j in range(m):
            for i in range(n):
                # Bits that need to be set via Column OR
                C[j] |= (B[i][j] & (~A[i][j]))

        R = [0] * n  # Row OR masks
        for i in range(n):
            for j in range(m):
                R[i] |= B[i][j]

        possible = True
        for i in range(n):
            for j in range(m):
                transformed = (A[i][j] | C[j]) & R[i]
                if transformed != B[i][j]:
                    possible = False
                    break
            if not possible:
                break

        print("Yes" if possible else "No")

threading.Thread(target=main).start()

**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
3 3 3
3 3 3
3 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
import sys

def II() -> int : return int(input())
def MI() -> int : return map(int, input().split())
def TI() -> tuple[int] : return tuple(MI())
def LI() -> list[int] : return list(MI())
#str-input
def SI() -> str : return input()
def MSI() -> str : return input().split()
def SI_L() -> list[str] : return list(SI())
def SI_LI() -> list[int] : return list(map(int, SI()))
#multiple-input
def LLI(n) -> list[list[int]]: return [LI() for _ in range(n)]
def LSI(n) -> list[str]: return [SI() for _ in range(n)]
#1-indexを0-indexでinput
def MI_1() -> int : return map(lambda x:int(x)-1, input().split())
def TI_1() -> tuple[int] : return tuple(MI_1())
def LI_1() -> list[int] : return list(MI_1())

from bisect import bisect_left

def solve(a,b):
    col = set(range(0,n))
    row = set(range(0,m))

    ccol = [0]*n #残っている1
    crow = [0]*m #残っている0
    # もし0になったら消去開始
    for i in range(n):
        for j in range(m):
            if b[i][j]:
                ccol[i] += 1
            else:
                crow[j] += 1

    que = []
    for i in range(n):
        if ccol[i] == 0:
            col.discard(i)
            que.append((0,i))

    for j in range(m):
        if crow[j] == 0:
            row.discard(j)
            que.append((1,j))

    while que:
        t,x = que.pop()
        dis = []
        if t == 0:
            for p in row:
                crow[p] -= 1
                if crow[p] == 0:
                    dis.append(p)
                    que.append((1,p))
            for p in dis:
                row.discard(p)

        else:
            for p in col:
                if ccol[p] == 0:
                    dis.append(p)
                    que.append((0,p))
            for p in dis:
                col.discard(p)

    for i in col:
        for j in row:
            if a[i][j] != b[i][j]:
                return False
    return True

t = II()
for _ in range(t):
    n,m = MI()
    #n*m <= 1000

    a = LLI(n)
    b = LLI(n)

    for d in range(30):
        ra = [[0]*m for i in range(n)]
        rb = [[0]*m for i in range(n)]
        for i in range(n):
            for j in range(m):
                ra[i][j] = (a[i][j]>>d) & 1
                rb[i][j] = (b[i][j]>>d) & 1
        if not solve(ra,rb):
            print("No")
            break
    else:
        print("Yes")

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

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

Think step by step.

Response:
```plaintext
1
2 2
1 1
1 1
0 1
1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
import java.util.*;

public class Main extends U {

    int n, m;
    boolean[][] graph;
    boolean[] vis, pre;

    void solve() {
        n = sc.nextInt();
        m = sc.nextInt();
        int[][] a = new int[n][m], b = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                a[i][j] = sc.nextInt();
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                b[i][j] = sc.nextInt();
            }
        }
        int[][] ta = new int[n][m], tb = new int[n][m];
        graph = new boolean[n + m][];
        vis = new boolean[n + m];
        pre = new boolean[n + m];
        for (int i = 0; i < n; i++) {
            graph[i] = new boolean[m];
        }
        for (int j = n; j < n + m; j++) {
            graph[j] = new boolean[n];
        }
        boolean[] use = new boolean[n + m];
        for (int k = 0; k < 30; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    ta[i][j] = a[i][j] >> k & 1;
                    tb[i][j] = b[i][j] >> k & 1;
                }
            }
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    graph[i][j] = tb[i][j] == 1;
                }
            }
            for (int j = 0; j < m; j++) {
                for (int i = 0; i < n; i++) {
                    graph[j + n][i] = tb[i][j] == 0;
                }
            }
            Arrays.fill(use, false);
            Arrays.fill(pre, false);
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (ta[i][j] != tb[i][j]) {
                        if (tb[i][j] == 0) {
                            use[i] = true;
                        } else {
                            use[j + n] = true;
                        }
                    }
                }
            }
            for (int i = 0; i < n + m; i++) {
                if (use[i] && !pre[i]) {
                    if (dfs(i)) {
                        out.println("NO");
                        return;
                    }
                    for (int j = 0; j < n + m; j++) {
                        if (vis[j]) {
                            pre[j] = true;
                            vis[j] = false;
                        }
                    }
                }
            }
        }
        out.println("YES");
    }

    boolean dfs(int u) {
        if (pre[u]) {
            return false;
        }
        if (vis[u]) {
            return true;
        }
        vis[u] = true;
        if (graph[u].length == n) {
            for (int i = 0; i < n; i++) {
                if (graph[u][i] && dfs(i)) {
                    return true;
                }
            }
        } else {
            for (int i = 0; i < m; i++) {
                if (graph[u][i] && dfs(i + n)) {
                    return true;
                }
            }
        }
        return false;
    }

    public static void main(String[] args) {
        Main o = new Main();
        int t = o.sc.nextInt(); while (t-- > 0)
        o.solve();
        o.out.flush();
    }
}

class U {                                                                                FastReader sc=new FastReader();FastWriter out=new FastWriter();class FastReader{private java.io.InputStream is=System.in;private byte[] inbuf=new byte[1024];private char[] str=new char[16];private int lenbuf,ptrbuf,b;int readByte(){if(ptrbuf==lenbuf){ptrbuf=0;try{lenbuf=is.read(inbuf);}catch(Exception e){}if(lenbuf<=0)return-1;}return inbuf[ptrbuf++];}String next(){while((b=readByte())<33);int i=0;while(b>32){if(i==str.length)str=Arrays.copyOf(str,str.length<<1);str[i++]=(char)b;b=readByte();}return new String(str,0,i);}int nextInt(){return(int)nextLong();}long nextLong(){while((b=readByte())<33);boolean minus=b=='-';long num=minus?0:b-'0';while((b=readByte())>32)num=num*10+(b-'0');return minus?-num:num;}double nextDouble(){return Double.parseDouble(next());}}class FastWriter{private java.io.OutputStream out=System.out;private int tr=0,BUF_SIZE=8192;private byte[] buf=new byte[BUF_SIZE];private int countDigits(int v){return v>=100000?v>=10000000?v>=100000000?v>=1000000000?10:9:8:v>=1000000?7:6:v>=1000?v>=10000?5:4:v>=100?3:v>=10?2:1;}private int countDigits(long v){return v>=10000000000L?10+countDigits((int)(v/10000000000L)):v>=1000000000?10:countDigits((int)v);}private FastWriter write(byte b){buf[tr++]=b;if(tr==BUF_SIZE)innerflush();return this;}private FastWriter write(char c){return write((byte)c);}private FastWriter write(int x){if(x==Integer.MIN_VALUE){return write((long)x);}if(tr+12>=BUF_SIZE)innerflush();if(x<0){write((byte)'-');x=-x;}int d=countDigits(x);for(int i=tr+d-1;i>=tr;i--){buf[i]=(byte)('0'+x%10);x/=10;}tr+=d;return this;}private FastWriter write(long x){if(x==Long.MIN_VALUE){return write(""+x);}if(tr+21>=BUF_SIZE)innerflush();if(x<0){write((byte)'-');x=-x;}int d=countDigits(x);for(int i=tr+d-1;i>=tr;i--){buf[i]=(byte)('0'+x%10);x/=10;}tr+=d;return this;}private FastWriter write(double x,int precision){if(x<0){write('-');x=-x;}x+=Math.pow(10,-precision)/2;write((long)x).write(".");x-=(long)x;for(int i=0;i<precision;i++){x*=10;write((char)('0'+(int)x));x-=(int)x;}return this;}private FastWriter write(String s){for(int i=0;i<s.length();i++){buf[tr++]=(byte)s.charAt(i);if(tr==BUF_SIZE)innerflush();}return this;}void print(char c){write(c);}void print(String s){write(s);}void print(int x){write(x);}void print(long x){write(x);}void print(double x,int precision){write(x,precision);}void writeln(){write((byte)'\n');}void println(char c){write(c).writeln();}void println(int x){write(x).writeln();}void println(long x){write(x).writeln();}void println(double x,int precision){write(x,precision).writeln();}void println(String s){write(s).writeln();}void println(Object o){deepPrint(o,true);}private void deepPrint(Object o,boolean f){if(o==null){write(f?"null\n":"null");return;}Class<?>c=o.getClass();if(c.isArray()){if(o instanceof Object[]){Object[]t=(Object[])o;int n=t.length;boolean b=f&&n>0&&t[0]!=null&&(t[0].getClass().isArray()||t[0]instanceof Collection||t[0]instanceof Map);write(b?"[\n":"[");for(int i=0;i<n;i++){deepPrint(t[i],false);if(i<n-1)write(b?", \n":", ");else if(b)writeln();}write("]");}else{write(o instanceof byte[]?Arrays.toString((byte[])o):o instanceof short[]?Arrays.toString((short[])o):o instanceof int[]?Arrays.toString((int[])o):o instanceof long[]?Arrays.toString((long[])o):o instanceof char[]?Arrays.toString((char[])o):o instanceof float[]?Arrays.toString((float[])o):o instanceof double[]?Arrays.toString((double[])o):Arrays.toString((boolean[])o));}}else if(o instanceof Collection){Collection<?>t=(Collection<?>)o;write("[");int i=0;for(Object v:t){deepPrint(v,false);if(i<t.size()-1)write(", ");i++;}write("]");}else if(o instanceof Map){Map<?,?>t=(Map<?,?>)o;write(!t.isEmpty()&&f?"{\n":"{");int i=0;for(Map.Entry<?,?>v:t.entrySet()){deepPrint(v.getKey(),false);write(" = ");deepPrint(v.getValue(),false);if(++i<t.size())write(f?", \n":", ");else if(f)writeln();}write("}");}else{write(o.toString());}if(f)writeln();}private void innerflush(){try{out.write(buf,0,tr);tr=0;}catch(Exception e){}}void flush(){innerflush();try{out.flush();}catch(Exception e){}}}int min(int a,int b){return a>b?b:a;}int min(int...args){int ans=Integer.MAX_VALUE;for(int v:args)if(ans>v)ans=v;return ans;}long min(long a,long b){return a>b?b:a;}long min(long...args){long ans=Long.MAX_VALUE;for(long v:args)if(ans>v)ans=v;return ans;}double min(double a,double b){return a>b?b:a;}double min(double...args){double ans=Double.MAX_VALUE;for(double v:args)if(ans>v)ans=v;return ans;}int max(int a,int b){return a<b?b:a;}int max(int...args){int ans=Integer.MIN_VALUE;for(int v:args)if(ans<v)ans=v;return ans;}long max(long a,long b){return a<b?b:a;}long max(long...args){long ans=Long.MIN_VALUE;for(long v:args)if(ans<v)ans=v;return ans;}double max(double a,double b){return a<b?b:a;}double max(double...args){double ans=Double.MIN_VALUE;for(double v:args)if(ans<v)ans=v;return ans;}int abs(int a){return a<0?-a:a;}long abs(long a){return a<0?-a:a;}double abs(double a){return a<0?-a:a;}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider the following aspects:

1. **Edge Cases**: Test cases with minimal or maximal values.
2. **Special Patterns**: Test cases with specific patterns that might cause the algorithm to fail.
3. **Non-Transformable Cases**: Test cases where the transformation is not possible.

Given the constraints and the nature of the problem, a small-scale test input that is complete and valid can be designed to expose the bug. Let's consider a test case where the transformation is not possible due to the bitwise operations.

### Challenging Test Input

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

### Explanation

1. **Matrix A**:
   ```
   1 0
   0 1
   ```

2. **Matrix B**:
   ```
   0 1
   1 0
   ```

### Why This Test Input is Challenging

- **Bitwise AND Operation**: The AND operation can only reduce the value of elements in a row.
- **Bitwise OR Operation**: The OR operation can only increase the value of elements in a column.

In this test case:
- The element at position (1,1) in matrix A is 1, and in matrix B it is 0. To transform 1 to 0, we need to apply an AND operation with 0, which would also affect the element at (1,2), making it 0. This would make the entire row 0, which is not the desired transformation.
- The element at position (2,2) in matrix A is 1, and in matrix B it is 0. Similarly, applying an AND operation with 0 would affect the element at (2,1), making it 0, which is not the desired transformation.

Thus, it is impossible to transform matrix A into matrix B using the given operations. This test input should expose the bug in the buggy code by forcing it to recognize that the transformation is not possible.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Matrix Transformation
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given two matrices $$$A$$$ and $$$B$$$ of size $$$n \times m$$$, filled with integers between $$$0$$$ and $$$10^9$$$. You can perform the following operations on matrix $$$A$$$ in any order and any number of times:

- &=: choose two integers $$$i$$$ and $$$x$$$ ($$$1 \le i \le n$$$, $$$x \ge 0$$$) and replace each element in row $$$i$$$ with the result of the bitwise AND operation between $$$x$$$ and that element. Formally, for every $$$j \in [1, m]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ & } x$$$;
- |=: choose two integers $$$j$$$ and $$$x$$$ ($$$1 \le j \le m$$$, $$$x \ge 0$$$) and replace each element in column $$$j$$$ with the result of the bitwise OR operation between $$$x$$$ and that element. Formally, for every $$$i \in [1, n]$$$, the element $$$A_{i,j}$$$ is replaced with $$$A_{i,j} \text{ | } x$$$.

The value of $$$x$$$ may be chosen differently for different operations.

Determine whether it is possible to transform matrix $$$A$$$ into matrix $$$B$$$ using the given operations any number of times (including zero).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. Then, $$$t$$$ test cases follow.

Each test case is given as follows:

- the first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 10^3$$$; $$$n \cdot m \le 10^3$$$) — the dimensions of the matrices $$$A$$$ and $$$B$$$;
- the following $$$n$$$ lines describe the matrix $$$A$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$A_{i,1}, A_{i,2}, \dots, A_{i,m}$$$ ($$$0 \le A_{i,j} \le 10^9$$$);
- the following $$$n$$$ lines describe the matrix $$$B$$$, where the $$$i$$$-th line contains $$$m$$$ integers $$$B_{i,1}, B_{i,2}, \dots, B_{i,m}$$$ ($$$0 \le B_{i,j} \le 10^9$$$).

Output: For each test case, output Yes if it is possible to transform the matrix $$$A$$$ into the matrix $$$B$$$; otherwise, output No. Each letter can be output in any case, upper or lower.

Examples:
input:
4
1 1
12
13
2 2
10 10
42 42
21 21
21 21
2 2
74 10
42 106
21 85
85 21
2 4
1 2 3 4
5 6 7 8
3 2 3 4
1 0 1 0
output:
Yes
Yes
No
Yes

Note: Let's consider the second set of input data and show a sequence of operations that transforms matrix $$$A$$$ into matrix $$$B$$$:

Initially, the matrix looks like this:

$$$\begin{bmatrix} 10&10\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 1$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 42&42\\ \end{bmatrix}$$$

Apply an operation of the first type with parameters $$$i = 2$$$ and $$$x = 0$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 0&0\\ 0&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 1$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&0\\ 21&0\\ \end{bmatrix}$$$

Apply an operation of the second type with parameters $$$j = 2$$$ and $$$x = 21$$$. As a result, we get the matrix:

$$$\begin{bmatrix} 21&21\\ 21&21\\ \end{bmatrix}$$$

Thus, we have transformed matrix $$$A$$$ into matrix $$$B$$$.

Buggy Code:
import java.util.*;

public class Main extends U {

/*

1
2 2
1 0
0 1
0 1
1 0

 */

    void solve() {
        int n = sc.nextInt(), m = sc.nextInt();
        int[][] a = new int[n][m], b = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                a[i][j] = sc.nextInt();
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                b[i][j] = sc.nextInt();
            }
        }
        int[][] ta = new int[n][m], tb = new int[n][m];
        boolean[] row = new boolean[n], col = new boolean[m];
        boolean[][] graph = new boolean[n + m][];
        int[] in = new int[n + m];
        for (int i = 0; i < n; i++) {
            graph[i] = new boolean[m];
        }
        for (int j = n; j < n + m; j++) {
            graph[j] = new boolean[n];
        }
        for (int k = 0; k < 30; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    ta[i][j] = a[i][j] >> k & 1;
                    tb[i][j] = b[i][j] >> k & 1;
                }
            }
            Arrays.fill(row, false);
            Arrays.fill(col, false);
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    int t = col[j] ? 1 : ta[i][j];
                    if (t != tb[i][j]) {
                        if (t == 1) {
                            row[i] = true;
                            break;
                        }
                    }
                }
                for (int j = 0; j < m; j++) {
                    if (row[i] && col[j]) {
                        continue;
                    }
                    if (!col[j] && (row[i] ? 0 : ta[i][j]) != tb[i][j]) {
                        col[j] = true;
                    }
                }
            }
            for (int i = 0; i < graph.length; i++) {
                Arrays.fill(graph[i], false);
            }
            Arrays.fill(in, 0);
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (row[i] && col[j]) {
                        if (tb[i][j] == 0) {
                            graph[j + n][i] = true;
                            in[i]++;
                        } else {
                            graph[i][j] = true;
                            in[j]++;
                        }
                    }
                }
            }
            Deque<Integer> queue = new ArrayDeque<>();
            for (int i = 0; i < graph.length; i++) {
                if (in[i] == 0) {
                    queue.addLast(i);
                }
            }
            int rest = n + m;
            while (!queue.isEmpty()) {
                int u = queue.removeFirst();
                rest--;
                if (graph[u].length == n) {
                    for (int i = 0; i < n; i++) {
                        if (graph[u][i] && --in[i] == 0) {
                            queue.addLast(i);
                        }
                    }
                } else {
                    for (int i = 0; i < m; i++) {
                        if (graph[u][i] && --in[i + n] == 0) {
                            queue.addLast(i + n);
                        }
                    }
                }
            }
            if (rest > 0) {
                out.println("NO");
                return;
            }
        }
        out.println("YES");
    }

    public static void main(String[] args) {
        Main o = new Main();
        int t = o.sc.nextInt(); while (t-- > 0)
        o.solve();
        o.out.flush();
    }
}

class U {                                                                                FastReader sc=new FastReader();FastWriter out=new FastWriter();class FastReader{private java.io.InputStream is=System.in;private byte[] inbuf=new byte[1024];private char[] str=new char[16];private int lenbuf,ptrbuf,b;int readByte(){if(ptrbuf==lenbuf){ptrbuf=0;try{lenbuf=is.read(inbuf);}catch(Exception e){}if(lenbuf<=0)return-1;}return inbuf[ptrbuf++];}String next(){while((b=readByte())<33);int i=0;while(b>32){if(i==str.length)str=Arrays.copyOf(str,str.length<<1);str[i++]=(char)b;b=readByte();}return new String(str,0,i);}int nextInt(){return(int)nextLong();}long nextLong(){while((b=readByte())<33);boolean minus=b=='-';long num=minus?0:b-'0';while((b=readByte())>32)num=num*10+(b-'0');return minus?-num:num;}double nextDouble(){return Double.parseDouble(next());}}class FastWriter{private java.io.OutputStream out=System.out;private int tr=0,BUF_SIZE=8192;private byte[] buf=new byte[BUF_SIZE];private int countDigits(int v){return v>=100000?v>=10000000?v>=100000000?v>=1000000000?10:9:8:v>=1000000?7:6:v>=1000?v>=10000?5:4:v>=100?3:v>=10?2:1;}private int countDigits(long v){return v>=10000000000L?10+countDigits((int)(v/10000000000L)):v>=1000000000?10:countDigits((int)v);}private FastWriter write(byte b){buf[tr++]=b;if(tr==BUF_SIZE)innerflush();return this;}private FastWriter write(char c){return write((byte)c);}private FastWriter write(int x){if(x==Integer.MIN_VALUE){return write((long)x);}if(tr+12>=BUF_SIZE)innerflush();if(x<0){write((byte)'-');x=-x;}int d=countDigits(x);for(int i=tr+d-1;i>=tr;i--){buf[i]=(byte)('0'+x%10);x/=10;}tr+=d;return this;}private FastWriter write(long x){if(x==Long.MIN_VALUE){return write(""+x);}if(tr+21>=BUF_SIZE)innerflush();if(x<0){write((byte)'-');x=-x;}int d=countDigits(x);for(int i=tr+d-1;i>=tr;i--){buf[i]=(byte)('0'+x%10);x/=10;}tr+=d;return this;}private FastWriter write(double x,int precision){if(x<0){write('-');x=-x;}x+=Math.pow(10,-precision)/2;write((long)x).write(".");x-=(long)x;for(int i=0;i<precision;i++){x*=10;write((char)('0'+(int)x));x-=(int)x;}return this;}private FastWriter write(String s){for(int i=0;i<s.length();i++){buf[tr++]=(byte)s.charAt(i);if(tr==BUF_SIZE)innerflush();}return this;}void print(char c){write(c);}void print(String s){write(s);}void print(int x){write(x);}void print(long x){write(x);}void print(double x,int precision){write(x,precision);}void writeln(){write((byte)'\n');}void println(char c){write(c).writeln();}void println(int x){write(x).writeln();}void println(long x){write(x).writeln();}void println(double x,int precision){write(x,precision).writeln();}void println(String s){write(s).writeln();}void println(Object o){deepPrint(o,true);}private void deepPrint(Object o,boolean f){if(o==null){write(f?"null\n":"null");return;}Class<?>c=o.getClass();if(c.isArray()){if(o instanceof Object[]){Object[]t=(Object[])o;int n=t.length;boolean b=f&&n>0&&t[0]!=null&&(t[0].getClass().isArray()||t[0]instanceof Collection||t[0]instanceof Map);write(b?"[\n":"[");for(int i=0;i<n;i++){deepPrint(t[i],false);if(i<n-1)write(b?", \n":", ");else if(b)writeln();}write("]");}else{write(o instanceof byte[]?Arrays.toString((byte[])o):o instanceof short[]?Arrays.toString((short[])o):o instanceof int[]?Arrays.toString((int[])o):o instanceof long[]?Arrays.toString((long[])o):o instanceof char[]?Arrays.toString((char[])o):o instanceof float[]?Arrays.toString((float[])o):o instanceof double[]?Arrays.toString((double[])o):Arrays.toString((boolean[])o));}}else if(o instanceof Collection){Collection<?>t=(Collection<?>)o;write("[");int i=0;for(Object v:t){deepPrint(v,false);if(i<t.size()-1)write(", ");i++;}write("]");}else if(o instanceof Map){Map<?,?>t=(Map<?,?>)o;write(!t.isEmpty()&&f?"{\n":"{");int i=0;for(Map.Entry<?,?>v:t.entrySet()){deepPrint(v.getKey(),false);write(" = ");deepPrint(v.getValue(),false);if(++i<t.size())write(f?", \n":", ");else if(f)writeln();}write("}");}else{write(o.toString());}if(f)writeln();}private void innerflush(){try{out.write(buf,0,tr);tr=0;}catch(Exception e){}}void flush(){innerflush();try{out.flush();}catch(Exception e){}}}int min(int a,int b){return a>b?b:a;}int min(int...args){int ans=Integer.MAX_VALUE;for(int v:args)if(ans>v)ans=v;return ans;}long min(long a,long b){return a>b?b:a;}long min(long...args){long ans=Long.MAX_VALUE;for(long v:args)if(ans>v)ans=v;return ans;}double min(double a,double b){return a>b?b:a;}double min(double...args){double ans=Double.MAX_VALUE;for(double v:args)if(ans>v)ans=v;return ans;}int max(int a,int b){return a<b?b:a;}int max(int...args){int ans=Integer.MIN_VALUE;for(int v:args)if(ans<v)ans=v;return ans;}long max(long a,long b){return a<b?b:a;}long max(long...args){long ans=Long.MIN_VALUE;for(long v:args)if(ans<v)ans=v;return ans;}double max(double a,double b){return a<b?b:a;}double max(double...args){double ans=Double.MIN_VALUE;for(double v:args)if(ans<v)ans=v;return ans;}int abs(int a){return a<0?-a:a;}long abs(long a){return a<0?-a:a;}double abs(double a){return a<0?-a:a;}
}

**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 that might cause the algorithm to fail. The key operations in the problem are bitwise AND and bitwise OR, and the goal is to transform matrix \(A\) into matrix \(B\).

### Step-by-Step Analysis:

1. **Edge Case 1: Zero Matrix Transformation**
   - If matrix \(A\) is all zeros and matrix \(B\) is also all zeros, the transformation is trivially possible.
   - If matrix \(A\) is all zeros and matrix \(B\) has any non-zero elements, the transformation is impossible.

2. **Edge Case 2: Single Element Matrices**
   - If both matrices are \(1 \times 1\) and the single element in \(A\) can be transformed into the single element in \(B\) using the allowed operations, the transformation is possible.
   - If the single element in \(A\) cannot be transformed into the single element in \(B\), the transformation is impossible.

3. **Edge Case 3: Bitwise Constraints**
   - If matrix \(A\) has elements that are all even and matrix \(B\) has elements that are all odd, the transformation is impossible because bitwise AND and OR operations cannot change the parity of all elements in the matrix.

4. **Edge Case 4: Mixed Elements**
   - If matrix \(A\) has a mix of even and odd elements and matrix \(B\) also has a mix of even and odd elements, the transformation might be possible, but it depends on the specific values and positions of the elements.

### Challenging Test Input:

Let's consider a test case where matrix \(A\) has a mix of even and odd elements, and matrix \(B\) has a mix of even and odd elements, but the transformation is impossible due to the bitwise constraints.

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

### Explanation:

- **Matrix \(A\):**
  - \[
    \begin{bmatrix}
    1 & 0 \\
    0 & 1 \\
    \end{bmatrix}
  \]

- **Matrix \(B\):**
  - \[
    \begin{bmatrix}
    0 & 1 \\
    1 & 0 \\
    \end{bmatrix}
  \]

### Why This Test Case is Challenging:

- **Bitwise AND Operations:**
  - Applying a bitwise AND operation with any \(x\) on a row or column will not change the parity of the elements. For example, if \(x\) is even, it will not change an odd element to even, and vice versa.

- **Bitwise OR Operations:**
  - Applying a bitwise OR operation with any \(x\) on a row or column will not change the parity of the elements. For example, if \(x\) is even, it will not change an odd element to even, and vice versa.

- **Transformation Impossibility:**
  - The elements in matrix \(A\) are \(1, 0, 0, 1\), which have mixed parity (odd and even).
  - The elements in matrix \(B\) are \(0, 1, 1, 0\), which also have mixed parity.
  - However, it is impossible to transform the elements of matrix \(A\) to match the elements of matrix \(B\) using the given operations because the parity of the elements cannot be changed in the required way.

This test case should expose the bug in the provided code by demonstrating a situation where the transformation is impossible due to the bitwise constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene's Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Nene invented a new game based on an increasing sequence of integers $$$a_1, a_2, \ldots, a_k$$$.

In this game, initially $$$n$$$ players are lined up in a row. In each of the rounds of this game, the following happens:

- Nene finds the $$$a_1$$$-th, $$$a_2$$$-th, $$$\ldots$$$, $$$a_k$$$-th players in a row. They are kicked out of the game simultaneously. If the $$$i$$$-th player in a row should be kicked out, but there are fewer than $$$i$$$ players in a row, they are skipped.

Once no one is kicked out of the game in some round, all the players that are still in the game are declared as winners.

For example, consider the game with $$$a=[3, 5]$$$ and $$$n=5$$$ players. Let the players be named player A, player B, $$$\ldots$$$, player E in the order they are lined up initially. Then,

- Before the first round, players are lined up as ABCDE. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. These are players C and E. They are kicked out in the first round.
- Now players are lined up as ABD. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. The $$$3$$$-rd player is player D and there is no $$$5$$$-th player in a row. Thus, only player D is kicked out in the second round.
- In the third round, no one is kicked out of the game, so the game ends after this round.
- Players A and B are declared as the winners.

Nene has not yet decided how many people would join the game initially. Nene gave you $$$q$$$ integers $$$n_1, n_2, \ldots, n_q$$$ and you should answer the following question for each $$$1 \le i \le q$$$ independently:

- How many people would be declared as winners if there are $$$n_i$$$ players in the game initially?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 250$$$). The description of test cases follows.

The first line case contains two integers $$$k$$$ and $$$q$$$ ($$$1 \le k, q \le 100$$$) — the length of the sequence $$$a$$$ and the number of values $$$n_i$$$ you should solve this problem for.

The second line contains $$$k$$$ integers $$$a_1,a_2,\ldots,a_k$$$ ($$$1\leq a_1<a_2<\ldots<a_k\leq 100$$$) — the sequence $$$a$$$.

The third line contains $$$q$$$ integers $$$n_1,n_2,\ldots,n_q$$$ ($$$1\leq n_i \leq 100$$$).

Output: For each test case, output $$$q$$$ integers: the $$$i$$$-th ($$$1\le i \le q$$$) of them should be the number of players declared as winners if initially $$$n_i$$$ players join the game.

Examples:
input:
6
2 1
3 5
5
5 3
2 4 6 7 9
1 3 5
5 4
3 4 5 6 7
1 2 3 4
2 3
69 96
1 10 100
1 1
100
50
3 3
10 20 30
1 10 100
output:
2
1 1 1
1 2 2 2
1 10 68
50
1 9 9

Note: The first test case was explained in the statement.

In the second test case, when $$$n=1$$$, the only player stays in the game in the first round. After that, the game ends and the only player is declared as a winner.

Buggy Code:
#include<iostream>
using namespace std;

int t, k, q, n, a[1001];

int bn_find(int l, int r, int x, int *a)
{
    int vt = -1;
    while (l <= r)
    {
        int mid = (l + r) / 2;
        if (a[mid] == x)
        {
            vt = mid;
            return vt;
        }
        else if (a[mid] < x)
        {
            vt = mid;
            l = mid + 1;
        }
        else
            r = mid - 1;
    }
    return vt;
}

void inp()
{
    cin >> k >> q;
    for (int i = 1; i <= k; i++)
        cin >> a[i];
}

void solve()
{
    cin >> t;
    while (t--)
    {
        inp();
        for (int i = 1; i <= q; i++)
        {
            cin >> n;
            if (n == 1)
                cout << 1 << " ";
            else
            {
                while (n > 1)
                {
                    int tmp = bn_find(1, k, n, a);
                    if (tmp != -1)
                        n -= tmp;
                    else
                        break;
                }
                cout << n << " ";
            }
        }
        cout << endl;
    }
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    solve();
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene's Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Nene invented a new game based on an increasing sequence of integers $$$a_1, a_2, \ldots, a_k$$$.

In this game, initially $$$n$$$ players are lined up in a row. In each of the rounds of this game, the following happens:

- Nene finds the $$$a_1$$$-th, $$$a_2$$$-th, $$$\ldots$$$, $$$a_k$$$-th players in a row. They are kicked out of the game simultaneously. If the $$$i$$$-th player in a row should be kicked out, but there are fewer than $$$i$$$ players in a row, they are skipped.

Once no one is kicked out of the game in some round, all the players that are still in the game are declared as winners.

For example, consider the game with $$$a=[3, 5]$$$ and $$$n=5$$$ players. Let the players be named player A, player B, $$$\ldots$$$, player E in the order they are lined up initially. Then,

- Before the first round, players are lined up as ABCDE. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. These are players C and E. They are kicked out in the first round.
- Now players are lined up as ABD. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. The $$$3$$$-rd player is player D and there is no $$$5$$$-th player in a row. Thus, only player D is kicked out in the second round.
- In the third round, no one is kicked out of the game, so the game ends after this round.
- Players A and B are declared as the winners.

Nene has not yet decided how many people would join the game initially. Nene gave you $$$q$$$ integers $$$n_1, n_2, \ldots, n_q$$$ and you should answer the following question for each $$$1 \le i \le q$$$ independently:

- How many people would be declared as winners if there are $$$n_i$$$ players in the game initially?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 250$$$). The description of test cases follows.

The first line case contains two integers $$$k$$$ and $$$q$$$ ($$$1 \le k, q \le 100$$$) — the length of the sequence $$$a$$$ and the number of values $$$n_i$$$ you should solve this problem for.

The second line contains $$$k$$$ integers $$$a_1,a_2,\ldots,a_k$$$ ($$$1\leq a_1<a_2<\ldots<a_k\leq 100$$$) — the sequence $$$a$$$.

The third line contains $$$q$$$ integers $$$n_1,n_2,\ldots,n_q$$$ ($$$1\leq n_i \leq 100$$$).

Output: For each test case, output $$$q$$$ integers: the $$$i$$$-th ($$$1\le i \le q$$$) of them should be the number of players declared as winners if initially $$$n_i$$$ players join the game.

Examples:
input:
6
2 1
3 5
5
5 3
2 4 6 7 9
1 3 5
5 4
3 4 5 6 7
1 2 3 4
2 3
69 96
1 10 100
1 1
100
50
3 3
10 20 30
1 10 100
output:
2
1 1 1
1 2 2 2
1 10 68
50
1 9 9

Note: The first test case was explained in the statement.

In the second test case, when $$$n=1$$$, the only player stays in the game in the first round. After that, the game ends and the only player is declared as a winner.

Buggy Code:
#include <iostream>

using namespace std;

int main()
{

   int t;
   cin>>t;
   while(t--)
   {
       int k,q;
       cin>>k>>q;
       int a[k],n[q];
       for(int i=0;i<k;i++)
        cin>>a[i];
         for(int i=0;i<q;i++)
        cin>>n[i];
        for(int i=0;i<q;i++)
        {
            if(n[i]>a[k-1])n[i]=a[0]-1;

           while(n[i]>=a[0])
           {
             int l=0,h=k-1,m=0;int ans=0;
             while(l<=h)
             {
                 m=(l+h)/2;
                 if(a[m]==n[i])
                    {ans=0;

                        n[i]=n[i]-m-1;;
                        break;
                    }
                else if(a[m]>n[i])
                      h=m-1;
                 else
                    {
                        ans=a[m];
                         l=m+1;
                    }
             }
             n[i]=n[i]-ans;
           }

        }
        for(int i=0;i<q;i++)
    cout<<n[i]<<" ";
    cout<<"\n";

   }

}

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

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

Think step by step.

Response:
```plaintext
3
2 2
3 5
5 6
5 3
2 4 6 7 9
1 3 5
2 3
69 96
1 10 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene's Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Nene invented a new game based on an increasing sequence of integers $$$a_1, a_2, \ldots, a_k$$$.

In this game, initially $$$n$$$ players are lined up in a row. In each of the rounds of this game, the following happens:

- Nene finds the $$$a_1$$$-th, $$$a_2$$$-th, $$$\ldots$$$, $$$a_k$$$-th players in a row. They are kicked out of the game simultaneously. If the $$$i$$$-th player in a row should be kicked out, but there are fewer than $$$i$$$ players in a row, they are skipped.

Once no one is kicked out of the game in some round, all the players that are still in the game are declared as winners.

For example, consider the game with $$$a=[3, 5]$$$ and $$$n=5$$$ players. Let the players be named player A, player B, $$$\ldots$$$, player E in the order they are lined up initially. Then,

- Before the first round, players are lined up as ABCDE. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. These are players C and E. They are kicked out in the first round.
- Now players are lined up as ABD. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. The $$$3$$$-rd player is player D and there is no $$$5$$$-th player in a row. Thus, only player D is kicked out in the second round.
- In the third round, no one is kicked out of the game, so the game ends after this round.
- Players A and B are declared as the winners.

Nene has not yet decided how many people would join the game initially. Nene gave you $$$q$$$ integers $$$n_1, n_2, \ldots, n_q$$$ and you should answer the following question for each $$$1 \le i \le q$$$ independently:

- How many people would be declared as winners if there are $$$n_i$$$ players in the game initially?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 250$$$). The description of test cases follows.

The first line case contains two integers $$$k$$$ and $$$q$$$ ($$$1 \le k, q \le 100$$$) — the length of the sequence $$$a$$$ and the number of values $$$n_i$$$ you should solve this problem for.

The second line contains $$$k$$$ integers $$$a_1,a_2,\ldots,a_k$$$ ($$$1\leq a_1<a_2<\ldots<a_k\leq 100$$$) — the sequence $$$a$$$.

The third line contains $$$q$$$ integers $$$n_1,n_2,\ldots,n_q$$$ ($$$1\leq n_i \leq 100$$$).

Output: For each test case, output $$$q$$$ integers: the $$$i$$$-th ($$$1\le i \le q$$$) of them should be the number of players declared as winners if initially $$$n_i$$$ players join the game.

Examples:
input:
6
2 1
3 5
5
5 3
2 4 6 7 9
1 3 5
5 4
3 4 5 6 7
1 2 3 4
2 3
69 96
1 10 100
1 1
100
50
3 3
10 20 30
1 10 100
output:
2
1 1 1
1 2 2 2
1 10 68
50
1 9 9

Note: The first test case was explained in the statement.

In the second test case, when $$$n=1$$$, the only player stays in the game in the first round. After that, the game ends and the only player is declared as a winner.

Buggy Code:
#include <iostream>
using namespace std;
int main()
{
    int t, m, key[100][100], q[100];
    cin >> t;
    for (int i = 0; i < t; i++) {
        m = 0;
        int a[100], n[100], k;
        cin >> k >> q[i];
        for (int p = 0; p < k; p++)
            cin >> a[p];
        for (int p = 0; p < q[i]; p++)
            cin >> n[p];
        while (m < q[i]) {
            if (n[m] < a[0])
                key[i][m] = n[m];
            else
                key[i][m] = a[0] - 1;
            m++;
        }
    }
    for (int i = 0; i < t; i++) {
        for (int p = 0; p < q[i]; p++)
            cout << key[i][p] << ' ';
        cout << 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:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following:

1. **Edge Cases**:
   - Very small values of `n` (e.g., 1, 2).
   - Values of `n` that are just larger than the largest `a_i` (e.g., if `a_k` is 100, test with `n = 101`).
   - Values of `n` that are multiples of the `a_i` values (e.g., if `a = [3, 5]`, test with `n = 15`).

2. **Complex Cases**:
   - Values of `n` where the sequence of `a` results in a complex elimination pattern.
   - Values of `n` where the sequence of `a` results in a small number of winners (e.g., if `a = [3, 5]`, test with `n = 6`).

3. **Large Values**:
   - Test with the maximum value of `n` (100) to ensure the code handles large inputs correctly.

4. **Multiple Test Cases**:
   - Include multiple test cases to cover different scenarios.

Based on these considerations, here is a small-scale test input that should expose the bug:

```plaintext
4
2 1
3 5
6
2 3
2 4 6 7 9
1 3 5
5 4
3 4 5 6 7
1 2 3 4
2 3
69 96
1 10 100
```

This input includes:
- A simple case with `n = 6` and `a = [3, 5]`.
- A case with `n = 1`, `n = 3`, and `n = 5` to test small values.
- A case with `n = 1`, `n = 2`, `n = 3`, and `n = 4` to test small values.
- A case with `n = 1`, `n = 10`, and `n = 100` to test a small, medium, and large value.

This should help identify the issues 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: Nene's Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Nene invented a new game based on an increasing sequence of integers $$$a_1, a_2, \ldots, a_k$$$.

In this game, initially $$$n$$$ players are lined up in a row. In each of the rounds of this game, the following happens:

- Nene finds the $$$a_1$$$-th, $$$a_2$$$-th, $$$\ldots$$$, $$$a_k$$$-th players in a row. They are kicked out of the game simultaneously. If the $$$i$$$-th player in a row should be kicked out, but there are fewer than $$$i$$$ players in a row, they are skipped.

Once no one is kicked out of the game in some round, all the players that are still in the game are declared as winners.

For example, consider the game with $$$a=[3, 5]$$$ and $$$n=5$$$ players. Let the players be named player A, player B, $$$\ldots$$$, player E in the order they are lined up initially. Then,

- Before the first round, players are lined up as ABCDE. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. These are players C and E. They are kicked out in the first round.
- Now players are lined up as ABD. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. The $$$3$$$-rd player is player D and there is no $$$5$$$-th player in a row. Thus, only player D is kicked out in the second round.
- In the third round, no one is kicked out of the game, so the game ends after this round.
- Players A and B are declared as the winners.

Nene has not yet decided how many people would join the game initially. Nene gave you $$$q$$$ integers $$$n_1, n_2, \ldots, n_q$$$ and you should answer the following question for each $$$1 \le i \le q$$$ independently:

- How many people would be declared as winners if there are $$$n_i$$$ players in the game initially?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 250$$$). The description of test cases follows.

The first line case contains two integers $$$k$$$ and $$$q$$$ ($$$1 \le k, q \le 100$$$) — the length of the sequence $$$a$$$ and the number of values $$$n_i$$$ you should solve this problem for.

The second line contains $$$k$$$ integers $$$a_1,a_2,\ldots,a_k$$$ ($$$1\leq a_1<a_2<\ldots<a_k\leq 100$$$) — the sequence $$$a$$$.

The third line contains $$$q$$$ integers $$$n_1,n_2,\ldots,n_q$$$ ($$$1\leq n_i \leq 100$$$).

Output: For each test case, output $$$q$$$ integers: the $$$i$$$-th ($$$1\le i \le q$$$) of them should be the number of players declared as winners if initially $$$n_i$$$ players join the game.

Examples:
input:
6
2 1
3 5
5
5 3
2 4 6 7 9
1 3 5
5 4
3 4 5 6 7
1 2 3 4
2 3
69 96
1 10 100
1 1
100
50
3 3
10 20 30
1 10 100
output:
2
1 1 1
1 2 2 2
1 10 68
50
1 9 9

Note: The first test case was explained in the statement.

In the second test case, when $$$n=1$$$, the only player stays in the game in the first round. After that, the game ends and the only player is declared as a winner.

Buggy Code:
t=int(input())
for i in range(t) :
    a,b = map(int, input().split())
    l=input().split(" ")
    l2=input().split(" ")
    l3=[]
    if(l[0]==1 and len(l)==1):
        for j in l2:
            l3.append("0")
    elif(l[0]=="1" and len(l)!=1):
        l3.append(l[0])
    else:
        for j in l2:
            if int(j)<int(l[0]):
                l3.append(j)
            else:
                l3.append(str(int(l[0])-1))
    print(" ".join(l3))

**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
2 4 6
1 2 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene's Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Nene invented a new game based on an increasing sequence of integers $$$a_1, a_2, \ldots, a_k$$$.

In this game, initially $$$n$$$ players are lined up in a row. In each of the rounds of this game, the following happens:

- Nene finds the $$$a_1$$$-th, $$$a_2$$$-th, $$$\ldots$$$, $$$a_k$$$-th players in a row. They are kicked out of the game simultaneously. If the $$$i$$$-th player in a row should be kicked out, but there are fewer than $$$i$$$ players in a row, they are skipped.

Once no one is kicked out of the game in some round, all the players that are still in the game are declared as winners.

For example, consider the game with $$$a=[3, 5]$$$ and $$$n=5$$$ players. Let the players be named player A, player B, $$$\ldots$$$, player E in the order they are lined up initially. Then,

- Before the first round, players are lined up as ABCDE. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. These are players C and E. They are kicked out in the first round.
- Now players are lined up as ABD. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. The $$$3$$$-rd player is player D and there is no $$$5$$$-th player in a row. Thus, only player D is kicked out in the second round.
- In the third round, no one is kicked out of the game, so the game ends after this round.
- Players A and B are declared as the winners.

Nene has not yet decided how many people would join the game initially. Nene gave you $$$q$$$ integers $$$n_1, n_2, \ldots, n_q$$$ and you should answer the following question for each $$$1 \le i \le q$$$ independently:

- How many people would be declared as winners if there are $$$n_i$$$ players in the game initially?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 250$$$). The description of test cases follows.

The first line case contains two integers $$$k$$$ and $$$q$$$ ($$$1 \le k, q \le 100$$$) — the length of the sequence $$$a$$$ and the number of values $$$n_i$$$ you should solve this problem for.

The second line contains $$$k$$$ integers $$$a_1,a_2,\ldots,a_k$$$ ($$$1\leq a_1<a_2<\ldots<a_k\leq 100$$$) — the sequence $$$a$$$.

The third line contains $$$q$$$ integers $$$n_1,n_2,\ldots,n_q$$$ ($$$1\leq n_i \leq 100$$$).

Output: For each test case, output $$$q$$$ integers: the $$$i$$$-th ($$$1\le i \le q$$$) of them should be the number of players declared as winners if initially $$$n_i$$$ players join the game.

Examples:
input:
6
2 1
3 5
5
5 3
2 4 6 7 9
1 3 5
5 4
3 4 5 6 7
1 2 3 4
2 3
69 96
1 10 100
1 1
100
50
3 3
10 20 30
1 10 100
output:
2
1 1 1
1 2 2 2
1 10 68
50
1 9 9

Note: The first test case was explained in the statement.

In the second test case, when $$$n=1$$$, the only player stays in the game in the first round. After that, the game ends and the only player is declared as a winner.

Buggy Code:
t=int(input())
for i in range(t) :
    a,b = map(int, input().split())
    l=input().split(" ")
    l2=input().split(" ")
    l3=[]
    if(l[0]=="1"):
        l3.append(l[0])
    else:
        for j in l2:
            if int(j)<int(l[0]):
                l3.append(j)
            else:
                l3.append(str(int(l[0])-1))
    print(" ".join(l3))

**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 3
3 5
1 3 5
5 3
1 2 3 4 5
1 100 100
3 1
10 20 30
50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene's Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Nene invented a new game based on an increasing sequence of integers $$$a_1, a_2, \ldots, a_k$$$.

In this game, initially $$$n$$$ players are lined up in a row. In each of the rounds of this game, the following happens:

- Nene finds the $$$a_1$$$-th, $$$a_2$$$-th, $$$\ldots$$$, $$$a_k$$$-th players in a row. They are kicked out of the game simultaneously. If the $$$i$$$-th player in a row should be kicked out, but there are fewer than $$$i$$$ players in a row, they are skipped.

Once no one is kicked out of the game in some round, all the players that are still in the game are declared as winners.

For example, consider the game with $$$a=[3, 5]$$$ and $$$n=5$$$ players. Let the players be named player A, player B, $$$\ldots$$$, player E in the order they are lined up initially. Then,

- Before the first round, players are lined up as ABCDE. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. These are players C and E. They are kicked out in the first round.
- Now players are lined up as ABD. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. The $$$3$$$-rd player is player D and there is no $$$5$$$-th player in a row. Thus, only player D is kicked out in the second round.
- In the third round, no one is kicked out of the game, so the game ends after this round.
- Players A and B are declared as the winners.

Nene has not yet decided how many people would join the game initially. Nene gave you $$$q$$$ integers $$$n_1, n_2, \ldots, n_q$$$ and you should answer the following question for each $$$1 \le i \le q$$$ independently:

- How many people would be declared as winners if there are $$$n_i$$$ players in the game initially?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 250$$$). The description of test cases follows.

The first line case contains two integers $$$k$$$ and $$$q$$$ ($$$1 \le k, q \le 100$$$) — the length of the sequence $$$a$$$ and the number of values $$$n_i$$$ you should solve this problem for.

The second line contains $$$k$$$ integers $$$a_1,a_2,\ldots,a_k$$$ ($$$1\leq a_1<a_2<\ldots<a_k\leq 100$$$) — the sequence $$$a$$$.

The third line contains $$$q$$$ integers $$$n_1,n_2,\ldots,n_q$$$ ($$$1\leq n_i \leq 100$$$).

Output: For each test case, output $$$q$$$ integers: the $$$i$$$-th ($$$1\le i \le q$$$) of them should be the number of players declared as winners if initially $$$n_i$$$ players join the game.

Examples:
input:
6
2 1
3 5
5
5 3
2 4 6 7 9
1 3 5
5 4
3 4 5 6 7
1 2 3 4
2 3
69 96
1 10 100
1 1
100
50
3 3
10 20 30
1 10 100
output:
2
1 1 1
1 2 2 2
1 10 68
50
1 9 9

Note: The first test case was explained in the statement.

In the second test case, when $$$n=1$$$, the only player stays in the game in the first round. After that, the game ends and the only player is declared as a winner.

Buggy Code:
m=int(input())
emp=[]
for i in range (m):
    k,q=map(int,input().split())
    a=sorted(list(map(int,input().split())))
    b=sorted(list(map(int,input().split())))
    for i in b:
        print(min(a[0]-1,i),end=" ")
    print()

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code's logic might fail. The buggy code always prints `min(a[0]-1, i)` for each `i` in the list of `b`, which is clearly incorrect. The correct logic should consider the actual game rules and simulate the rounds correctly.

### Key Points to Consider:
1. **Edge Cases:**
   - When `n` is less than `a[0]`.
   - When `n` is exactly `a[0]`.
   - When `n` is slightly larger than `a[0]`.
   - When `n` is much larger than `a[0]`.

2. **Scenarios with Multiple Rounds:**
   - Ensure the test case includes values of `n` that require multiple rounds to determine the winners.

3. **Small-Scale Test Input:**
   - The test input should be small enough to be manageable but complex enough to expose the bug.

### Chosen Test Input:
Let's create a test case with `t = 1` (one test case), `k = 3` (sequence length), and `q = 3` (number of `n` values to test).

### Test Case:
- `a = [3, 5, 7]`
- `n = [5, 10, 15]`

### Explanation:
- For `n = 5`:
  - Initial lineup: A B C D E
  - Round 1: Kick out 3rd and 5th players (C and E). Remaining: A B D
  - Round 2: Kick out 3rd player (D). Remaining: A B
  - Round 3: No one to kick out. Winners: A B (2 winners)

- For `n = 10`:
  - Initial lineup: A B C D E F G H I J
  - Round 1: Kick out 3rd, 5th, and 7th players (C, E, G). Remaining: A B D F H I J
  - Round 2: Kick out 3rd, 5th, and 7th players (F, H, J). Remaining: A B D I
  - Round 3: Kick out 3rd player (I). Remaining: A B D
  - Round 4: No one to kick out. Winners: A B D (3 winners)

- For `n = 15`:
  - Initial lineup: A B C D E F G H I J K L M N O
  - Round 1: Kick out 3rd, 5th, and 7th players (C, E, G). Remaining: A B D F H I J K L M N O
  - Round 2: Kick out 3rd, 5th, and 7th players (F, H, J). Remaining: A B D I K L M N O
  - Round 3: Kick out 3rd, 5th, and 7th players (L, N). Remaining: A B D I K M O
  - Round 4: Kick out 3rd player (M). Remaining: A B D I K O
  - Round 5: No one to kick out. Winners: A B D I K O (6 winners)

### Output:
```plaintext
1
3 3
3 5 7
5 10 15
```

This test input should effectively expose the bug in the provided buggy code by requiring multiple rounds of the game and involving different scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene's Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Nene invented a new game based on an increasing sequence of integers $$$a_1, a_2, \ldots, a_k$$$.

In this game, initially $$$n$$$ players are lined up in a row. In each of the rounds of this game, the following happens:

- Nene finds the $$$a_1$$$-th, $$$a_2$$$-th, $$$\ldots$$$, $$$a_k$$$-th players in a row. They are kicked out of the game simultaneously. If the $$$i$$$-th player in a row should be kicked out, but there are fewer than $$$i$$$ players in a row, they are skipped.

Once no one is kicked out of the game in some round, all the players that are still in the game are declared as winners.

For example, consider the game with $$$a=[3, 5]$$$ and $$$n=5$$$ players. Let the players be named player A, player B, $$$\ldots$$$, player E in the order they are lined up initially. Then,

- Before the first round, players are lined up as ABCDE. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. These are players C and E. They are kicked out in the first round.
- Now players are lined up as ABD. Nene finds the $$$3$$$-rd and the $$$5$$$-th players in a row. The $$$3$$$-rd player is player D and there is no $$$5$$$-th player in a row. Thus, only player D is kicked out in the second round.
- In the third round, no one is kicked out of the game, so the game ends after this round.
- Players A and B are declared as the winners.

Nene has not yet decided how many people would join the game initially. Nene gave you $$$q$$$ integers $$$n_1, n_2, \ldots, n_q$$$ and you should answer the following question for each $$$1 \le i \le q$$$ independently:

- How many people would be declared as winners if there are $$$n_i$$$ players in the game initially?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 250$$$). The description of test cases follows.

The first line case contains two integers $$$k$$$ and $$$q$$$ ($$$1 \le k, q \le 100$$$) — the length of the sequence $$$a$$$ and the number of values $$$n_i$$$ you should solve this problem for.

The second line contains $$$k$$$ integers $$$a_1,a_2,\ldots,a_k$$$ ($$$1\leq a_1<a_2<\ldots<a_k\leq 100$$$) — the sequence $$$a$$$.

The third line contains $$$q$$$ integers $$$n_1,n_2,\ldots,n_q$$$ ($$$1\leq n_i \leq 100$$$).

Output: For each test case, output $$$q$$$ integers: the $$$i$$$-th ($$$1\le i \le q$$$) of them should be the number of players declared as winners if initially $$$n_i$$$ players join the game.

Examples:
input:
6
2 1
3 5
5
5 3
2 4 6 7 9
1 3 5
5 4
3 4 5 6 7
1 2 3 4
2 3
69 96
1 10 100
1 1
100
50
3 3
10 20 30
1 10 100
output:
2
1 1 1
1 2 2 2
1 10 68
50
1 9 9

Note: The first test case was explained in the statement.

In the second test case, when $$$n=1$$$, the only player stays in the game in the first round. After that, the game ends and the only player is declared as a winner.

Buggy Code:
import java.util.*;

public class _1956A {
    public static void solve(int arr[], int q[], int n) {
        for (int i = 0; i < n; i++) {
            if (q[i] < arr[0]) {
                System.out.print(q[i] + " ");
            } else {
                while (q[i] >= arr[0]) {
                    q[i]--;
                }
                System.out.print(q[i] + " ");
            }
        }
        System.out.println();
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t > 0) {
            int k = sc.nextInt();
            int n = sc.nextInt();

            int arr[] = new int[k];
            int q[] = new int[n];

            for (int i = 0; i < k; i++) {
                arr[i] = sc.nextInt();
            }

            for (int i = 0; i < n; i++) {
                q[i] = sc.nextInt();
            }

            Arrays.sort(q);
            Arrays.sort(arr);

            solve(arr, q, n);
            t--;
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on scenarios where the code's logic fails. The primary issue in the buggy code is the incorrect handling of the game rounds and the player elimination process. Specifically, the code incorrectly decrements the number of players without properly simulating the rounds and player elimination.

Here are the steps to generate a challenging test input:

1. **Edge Case: Few Players**: Test with a small number of players to see if the code correctly handles the case where no one is eliminated in the first round.
2. **Edge Case: Large Sequence**: Test with a large sequence of elimination positions to see if the code can handle multiple rounds correctly.
3. **Edge Case: Mixed Scenarios**: Test with a mix of small and large player counts to see if the code can handle different scenarios.

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

```plaintext
4
2 1
3 5
5
5 3
2 4 6 7 9
1 3 5
5 4
3 4 5 6 7
1 2 3 4
2 3
69 96
1 10 100
1 1
100
50
3 3
10 20 30
1 10 100
```

This input includes:
1. A test case with 5 players and elimination positions 3 and 5.
2. A test case with 5 elimination positions and 3 player counts.
3. A test case with 5 elimination positions and 4 player counts.
4. A test case with 2 elimination positions and 3 player counts.
5. A test case with 1 elimination position and 1 player count.
6. A test case with 3 elimination positions and 3 player counts.

This variety of test cases will help expose the bug in the faulty code.