**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

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

#define int long long
#define fastio ios_base::sync_with_stdio(0); cin.tie(0)
#define endl "\n"

const int N=1e6+5;
vector<vector<int>> divisors(N);
vector<int> Primes;

void Sieve()
{
    int i, j, n=divisors.size();

    for(i=1; i<n; i++)
    {
        for(j=i; j<n; j+=i) divisors[j].push_back(i);
        if(divisors[i].size()==2) Primes.push_back(i);
    }
}

bool pathExists(vector<vector<int>>& grid, int start)
{
    int i=0, j=0, n=grid.size(), m=grid[0].size();

    auto dp=grid;
    dp[0][0]=start;

    for(j=1; j<m; j++) dp[0][j]=__gcd(grid[0][j], dp[0][j-1]);
    for(i=1; i<n; i++) dp[i][0]=__gcd(grid[i][0], dp[i-1][0]);

    for(i=1; i<n; i++) for(j=1; j<m; j++)
    {
        dp[i][j]=max(__gcd(grid[i][j], dp[i-1][j]), __gcd(grid[i][j], dp[i][j-1]));
    }

    return (dp[n-1][m-1]==start);
}

void pre()
{
    fastio;

    Sieve();
}

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

    vector<vector<int>> grid(n, vector<int>(m)), dp=grid;
    for(auto &row: grid) for(auto &it: row) cin >> it;

    int g=__gcd(grid[0][0], grid[n-1][m-1]), ans=1;
    set<int> st;

    for(auto it: divisors[g])
    {
        if(it==1) continue;
        for(auto jt: st) if(it%jt==0) continue;

        if(pathExists(grid, it)) ans=it;
        else st.insert(it);
    }

    cout << ans;
}

signed main()
{
    pre();

    int tc, tt=1;
    cin >> tt;

    for(tc=1; tc<=tt; tc++)
    {
        solve(tc);
        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:
```plaintext
1
2 2
1 2
3 4
```

**Explanation:**
This test case is designed to expose a potential issue in the buggy code. The grid is small and simple, with a clear path from the top-left corner to the bottom-right corner. The maximum GCD along any path in this grid is 1, which is the GCD of the numbers 1, 2, 3, and 4.

The path from (1,1) to (2,2) can be either:
1. (1,1) -> (1,2) -> (2,2) with GCDs: gcd(1, 2) = 1, gcd(1, 4) = 1
2. (1,1) -> (2,1) -> (2,2) with GCDs: gcd(1, 3) = 1, gcd(1, 4) = 1

In both cases, the maximum GCD is 1. This test case should help identify any issues in the logic that calculates the maximum GCD, particularly in how it handles the initial values and the path calculations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

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

// debug below
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }

template <typename T> void __print(const priority_queue<T> &q) {
  priority_queue<T> temp = q;
  cerr << '[';
  while (!temp.empty()) {
    __print(temp.top());
    temp.pop();
    cerr << (temp.empty() ? "" : ",");
  }
  cerr << ']';
}

template <typename T> void __print(const queue<T> &q) {
  queue<T> temp = q;
  cerr << '[';
  while (!temp.empty()) {
    __print(temp.front());
    temp.pop();
    cerr << (temp.empty() ? "" : ",");
  }
  cerr << ']';
}

template <typename T, typename V, typename U>
void __print(const tuple<T, V, U> &x) {
  cerr << '{';
  __print(get<0>(x));
  cerr << ',';
  __print(get<1>(x));
  cerr << ',';
  __print(get<2>(x));
  cerr << '}';
}

template <typename T, typename V> void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T> void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x)
    cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V> void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v))
    cerr << ", ";
  _print(v...);
}

#ifndef ONLINE_JUDGE
#define dbg(x...)                                                              \
  cerr << "[" << #x << "] = [";                                                \
  _print(x)
#else
#define dbg(x...)
#endif
// for fast code
#define PB push_back
#define ALL(x) (x).begin(), (x).end()
#define sze(x) ((int)(x).size())
#define RALL(x) (x).rbegin(), (x).rend()
#define FOR(i, j, k, in)                                                       \
  for (auto i = (in > 0) ? j : k; (in > 0) ? (i < k) : (i > k); i += in)
#define RFOR(i, j, k, in)                                                      \
  for (auto i = j; (in > 0) ? (i >= k) : (i <= k); i -= in)
#define REVERSE(a) std::reverse(a.begin(), a.end());
#define SORT(a) std::sort(a.begin(), a.end());
#define RSORT(a) std::sort(a.rbegin(), a.rend());
#define V std::vector
#define P std::pair
#define UST std::unordered_set
#define ST std::set
#define MST std::multiset
#define UMST std::unordered_multiset
#define UMP std::unordered_map
#define MP std::map
#define UMMP std::unordered_multimap
#define MMP std::multimap
#define PRQ std::priority_queue
#define Q std::queue
#define DQ std::deque
#define S std::stack
#define fi first
#define se second
#define PI 3.14159265358979323846

const long long INF = 2000000000000000000LL;
const int d4i[4] = {-1, 0, 1, 0}, d4j[4] = {0, 1, 0, -1};
typedef long long I;
typedef long long LL;
typedef unsigned long long ULL;
// typedef int I;
typedef float F;
typedef char C;
typedef bool B;
typedef double DB;
typedef long double LDB;
typedef string STR;
// const I mod = 1e9 + 7;
const I mod = 998244353;

#define equal_db(x, y) (std::fabs((x) - (y)) < (1e-9))

int madd(int a, int b, int c = mod) {
  int res = a + b;
  return (res >= c ? res - c : res);
}
int mminus(int a, int b, int c = mod) {
  int res;
  if (abs(a - b) < c)
    res = a - b;
  else
    res = (a - b) % c;
  return (res < 0 ? res + c : res);
}
int mmul(int a, int b, int c = mod) {
  LL res = (LL)a * b;
  return (res >= c ? res % c : res);
}
int mmuln(int a, int b, int c = mod) {
  LL res = (LL)a * b;
  return ((res % c) + c) % c;
}
LL mmulmod(LL a, LL b, LL m = mod) {
  LL q = (LL)(((LDB)a * (LDB)b) / (LDB)m);
  LL r = a * b - q * m;
  if (r > m)
    r %= m;
  if (r < 0)
    r += m;
  return r;
}

const I maxn = 1e6 + 1, maxm = 2e5 + 1;
V<I> divs[maxn];
void sieve() {
    for(int i=1;i<=maxn;i++) {
        for(int j=i;j<maxn;j+=i) {
            divs[j].PB(i);
        }
    }
}
void dfs(I x, I y, I n, I m, V<V<I>>&vis, V<V<I>> &a, I div) {
    if(x==n||y==m||x<0||y<0||vis[x][y]||a[x][y]%div) return;
    vis[x][y]=1;
    dfs(x+1,y,n,m,vis,a,div);
    dfs(x,y+1,n,m,vis,a,div);
}

void solve()
{
  I n,m;cin>>n>>m;
  V<V<I>> a(n,V<I>(m));
  FOR(i,0,n,1) FOR(j,0,m,1) cin>>a[i][j];
    I ans=1;
  for(auto&div:divs[a[0][0]]) {
      V<V<I>> vis(n,V<I>(m,0));
      dfs(0,0,n,m,vis,a,div);
      if(vis[n-1][m-1]) ans=max(ans,div);
  }
    cout<<ans<<'\n';

}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
    sieve();
  int t;
  cin >> t;
  while (t--)
    solve();
  // cnt++,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 provided code, we need to consider the following:

1. **Edge Cases**: The code should handle small and large grids, and cases where the grid is filled with numbers that have common divisors or not.
2. **Common Divisors**: The code should correctly handle cases where the numbers in the grid have common divisors, especially when the GCD is not 1.
3. **Path Constraints**: The code should correctly handle the constraints of moving only down and to the right.

Given these considerations, a good test input would be a grid where the numbers have a common divisor that is not 1, and the GCD along the path from the top-left to the bottom-right corner is maximized.

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

```plaintext
1
3 3
12 6 9
3 15 18
6 12 18
```

Explanation:
- The grid is 3x3.
- The numbers in the grid have a common divisor of 3.
- The path from (0,0) to (2,2) will have numbers that have a GCD of 3.

This test input should expose any issues with the code's ability to correctly compute the GCD along the path.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

Buggy Code:
from sys import stdin
input=lambda :stdin.readline()[:-1]
from math import gcd

def path_exists(grid, n, m):
    dp = [[False] * m for _ in range(n)]
    dp[0][0] = grid[0][0] == 1

    for i in range(n):
        for j in range(m):
            if grid[i][j] == 1:
                if i > 0:
                    dp[i][j] |= dp[i-1][j]
                if j > 0:
                    dp[i][j] |= dp[i][j-1]

    return dp[n-1][m-1]

def solve():
    n,m = map(int,input().split())
    matrix = []
    for i in range(n):
        a = list(map(int,input().split()))
        matrix.append(a)
    g = gcd(matrix[0][0], matrix[n-1][m-1])
    divs = []
    for i in range(1, int(g**0.5) + 2):
        if g % i == 0:
            divs.append(i)
            divs.append(g // i)
    divs = list(set(divs))
    divs.sort()

    max_divisor = 1
    for x in divs:
        b = [[1 if matrix[i][j] % x == 0 else 0 for j in range(m)] for i in range(n)]
        if path_exists(b, n, m):
            max_divisor = x

    print(max_divisor)

for __ in range(int(input())):
    solve()

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

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

Think step by step.

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

1. **Path Existence Check**: The `path_exists` function checks if a path exists where all cells are 1. However, this function is not correctly implemented to check if a path exists with all cells being divisible by a given divisor. It should check if there is a path from the top-left to the bottom-right where all cells are divisible by the divisor.

2. **Divisibility Check**: The `solve` function generates a list of divisors and checks if a path exists for each divisor. The `path_exists` function should be modified to correctly handle this.

3. **Edge Cases**: We need to consider edge cases where the grid has very small or very large values, and where the grid is small but has a complex structure that makes it difficult to find a valid path.

Given these points, a small-scale test input that can expose the bug is:

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

Explanation:
- The grid is 3x3, which is small but has a simple structure.
- The numbers in the grid are such that the GCD of the path from (1,1) to (3,3) is 1, as there is no common divisor greater than 1 for the numbers in the grid.
- The `path_exists` function will incorrectly return `False` for any divisor other than 1, as it checks if the cells are 1, not if they are divisible by the divisor.

This input should help expose the bug in the `path_exists` function and highlight the need for a correct implementation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

Buggy Code:
# Author: S Mahesh Raju
# Username: maheshraju2020
# Date: 08-04-2024 22:06
from random import randint
from sys import stdin, stdout, setrecursionlimit
from math import gcd, ceil, sqrt, lcm, comb
from collections import Counter, deque, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heappush, heappop, heapify
from types import GeneratorType
def ii1(): return int(stdin.readline().strip())
def is1(): return stdin.readline().strip()
def iia(): return list(map(int, stdin.readline().strip().split()))
def isa(): return stdin.readline().strip().split()

class Key(int):
    def __init__(self, x):
        int.__init__(x)

    def __hash__(self):
        return super(Key, self).__hash__() ^ randint(1, 10 ** 9)

def getDivisors(n):
    div = set()
    for i in range(1, int(sqrt(n)) + 2):
        if n % i == 0:
            div.add(i)
            div.add(n//i)
    return sorted(list(div))

tc = ii1()
for _ in range(tc):
    n, m = iia()
    arr = []
    for i in range(n):
        arr.append(iia())
    div = getDivisors(gcd(arr[0][0], arr[-1][-1]))
    res = 1
    for cur in div:
        dp = [[0 for i in range(m)] for j in range(n)]
        dp[0][0] = 1
        for i in range(n):
            for j in range(m):
                if i == 0 and j == 0:
                    continue
                elif j == 0:
                    if dp[i-1][j] and arr[i][j]%cur == 0:
                        dp[i][j] = 1
                elif i == 0:
                    if dp[i][j-1] and arr[i][j] % cur == 0:
                        dp[i][j] = 1
                else:
                    if dp[i-1][j] and arr[i][j] % cur == 0:
                        dp[i][j] = 1
                    if dp[i][j-1] and arr[i][j] % cur == 0:
                        dp[i][j] = 1
        if dp[-1][-1]:
            res = max(res, cur)
    print(res)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

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

public class Main {
    public static class pair{
        char c;
        int index;
        String type="";
        pair(char c, int index){
            this.c = c;
            this.index = index;
        }
    }
    static FastReader sc = new FastReader();
    static int max=0;
    static int n;
    static int m;
    static ArrayList<ArrayList<HashSet<Integer>>> s = new ArrayList<>();
    // --------------------------------------start-------------------------------------------//
    public static void solve(int tt) {
        n=sc.nextInt();
        m=sc.nextInt();
        ArrayList<ArrayList<HashSet<Integer>>> hs = new ArrayList<>();
        for(int i=0; i<n; i++){
            ArrayList<HashSet<Integer>> hs2 = new ArrayList<>();
            for(int j=0; j<m; j++){
                int in=sc.nextInt();
                hs2.add(factors(in));
            }
            hs.add(hs2);
        }
        s=hs;
        max=0;
        recur(0,0,s.get(0).get(0));
        System.out.println(max);
    }
    public static void recur(int cn,int cm,HashSet<Integer> cur){
        if(cur.size()==1)return;
        boolean found=false;
        for(int i:cur){
            if(i>max){
                found=true;
                break;
            }
        }
        if(!found)return;
        HashSet<Integer> passOn1=new HashSet<Integer>();
        int localMax=0;
        for(int i:cur){
            if(s.get(cn).get(cm).contains(i)){
                passOn1.add(i);
                localMax=Math.max(i,localMax);
            }
        }
        if(localMax<=max)return;
        if(cn+1<n){
            recur(cn+1,cm,passOn1);
        }
        if(cm+1<m){
            recur(cn, cm+1,passOn1);
        }
        if(cn==n-1&&cm==m-1){
            max=Math.max(localMax,max);
        }
    }
    public static void main(String[] args) {
        int t= sc.nextInt();
        int i=1;
        // ss();
        while(t-->0){
            solve(i++);
        }
    }
    // public static void ss(){
    //     for(int i=1;i<=500;i++){
    //         System.out.println(2*i);
    //         int ret=solve(i);
    //         if(ret>2*i){
    //             break;
    //         }
    //     }
    // }

    // public static long getSum(int r){
    // long sum=0;
    // int pp=r;
    // while(pp>0){
    // sum+=fenwickTree[pp];
    // pp-=pp&-pp;
    // }
    // return sum;
    // }
    // public static void update(int k,long v){
    // // long value=fenwickTree[k];
    // while(k<=n){
    // fenwickTree[k]+=(v);
    // k+=k&-k;
    // }
    // }
    static long factorial(long n) {
        long result = 1;
        for (long i = 1; i <= n; i++) {
            result *= i;
            if (result > 998244353) {
                result -= 998244353;
            }
        }
        return result % 998244353;
    }

    static int modFact(int n, int p) {
        if (n >= p)
            return 0;

        int result = 1;
        for (int i = 1; i <= n; i++)
            result = (result * i) % p;

        return result;
    }

    // Function to calculate combination
    // static long combination(long n, long r) {
    // return factorial(n) / (factorial(r) * factorial(n - r));
    // }

    // public static long binarySearch(ArrayList<Long> arr, long target) {
    // int left = 0, right = arr.size() - 1;
    // while (left <= right) {
    // int mid = left + (right - left) / 2;
    // if (arr.get(mid) == target) {
    // return 0;
    // } else if (arr.get(mid) < target) {
    // left = mid + 1;
    // } else {
    // right = mid - 1;
    // }
    // }
    // if (left == 0) {
    // return Math.abs(target - arr.get(0));
    // } else if (left == arr.size()) {
    // return Math.abs(arr.get(arr.size() - 1) - target);
    // } else {
    // long diff1 = Math.abs(target - arr.get(left - 1));
    // long diff2 = Math.abs(arr.get(left) - target);
    // return Math.min(diff1, diff2);
    // }
    // }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(
                    new InputStreamReader(System.in));
        }

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                if (st.hasMoreTokens()) {
                    str = st.nextToken("\n");
                } else {
                    str = br.readLine();
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    // public static int maxFrequencyNumber(int[] arr) {
    // if (arr.length == 0)
    // return -1;
    // int maxFreq = 0;
    // int number = -1;
    // HashMap<Integer, Integer> map = new HashMap<>();

    // for (int j : arr) {
    // if (map.containsKey(j)) {
    // map.put(j, map.get(j) + 1);
    // } else {
    // map.put(j, 1);
    // }
    // }
    // // using set data structure
    // Set<Integer> keySet = map.keySet();
    // for (Integer i : keySet) {
    // if (map.get(i) > maxFreq) {
    // number = i;
    // maxFreq = map.get(i);
    // }
    // }
    // return number;
    // }

    // public static boolean isPrime(long n) {
    // if (n <= 1)
    // return false;

    // // Check if n=2 or n=3
    // if (n == 2 || n == 3)
    // return true;

    // // Check whether n is divisible by 2 or 3
    // if (n % 2 == 0 || n % 3 == 0)
    // return false;

    // // Check from 5 to square root of n
    // // Iterate i by (i+6)
    // for (int i = 5; i <= Math.sqrt(n); i = i + 6)
    // if (n % i == 0 || n % (i + 2) == 0)
    // return false;

    // return true;

    // }

    // public static long lcm(int x, int y) {
    // int i;
    // int a = Math.max(x, y);

    // for (i = a; i <= x * y; i = i + a) {
    // if (i % x == 0 && i % y == 0)
    // break;
    // }
    // return i;
    // }

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

    // static ArrayList<Long> getFactors(long num) {
    // ArrayList<Long> factors = new ArrayList<Long>();
    // int i;
    // boolean flag = false;

    // for (i = 1; i <= Math.sqrt(num); i++) {
    // if (num % i == 0)
    // factors.add((long) i);
    // if (i == num / i)
    // flag = true;
    // }
    // if (flag)
    // i -= 2;

    // // printing pairs
    // for (; i >= 1; i--) {
    // if (num % i == 0)
    // factors.add((long) (num / i));
    // }
    // return factors;
    // }

    // static int fact(int number) {
    // int f = 1;
    // int j = 1;
    // while (j <= number) {
    // f = f * j;
    // j++;
    // }
    // return f;
    // }

    public static int[] arrInput(int n) {
        int[] arr = new int[n];
        for (int i = 0; i < arr.length; i++) {
            arr[i] = sc.nextInt();
        }
        return arr;
    }

    public static void arrOutput(int[] arr) {
        StringBuilder out = new StringBuilder("");
        for (int i = 0; i < arr.length; i++) {
            out.append(arr[i] + " ");
        }
        System.out.println(out);
    }

    // public static boolean isInt(double d) {
    // return d % 1 == 0;
    // }

    // public static boolean isPowerOfTwo(int n) {
    // return (int) (Math.ceil((Math.log(n) / Math.log(2)))) == (int)
    // (Math.floor(((Math.log(n) / Math.log(2)))));
    // }
    public static HashSet<Integer> factors(int n){
        HashSet<Integer> h = new HashSet<Integer>();
        for(int i=1;i*i<=n;i++){
            if(n%i == 0){
                h.add(i);
                h.add(n/i);
            }
        }
        return h;
    }
    public static ArrayList<Integer> primeFactorization(int numbers) {
        int n = numbers;
        ArrayList<Integer> factors = new ArrayList<Integer>();
        for (int i = 2; i <= n / i; i++) {
            while (n % i == 0) {
                factors.add(i);
                n /= i;
            }
        }
        if (n > 1) {
            factors.add(n);
        }
        return factors;
    }

    public static int calculate(int l, int r, int segTree[], int n) {
        l += n - 1;
        r += n - 1;
        int ans = segTree[r];
        while (l <= r) {
            if (r % 2 == 0) {
                ans &= segTree[r];
                r--;
            }
            if (l % 2 == 1) {
                ans &= segTree[l];
                l++;
            }
            r = r >> 1;
            l = l >> 1;
        }
        return ans;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

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

public class GCDOnAGrid {
    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        int t = Integer.parseInt(br.readLine());
        for(int q = 0; q < t; q++){
            StringTokenizer st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            int m = Integer.parseInt(st.nextToken());
            int[][] grid = new int[n][m];
            ArrayList<HashSet<Integer>> been = new ArrayList<HashSet<Integer>>();
            for(int i = 0; i < n; i++){
                st = new StringTokenizer(br.readLine());
                for(int j = 0; j < m; j++){
                    grid[i][j] = Integer.parseInt(st.nextToken());
                    been.add(new HashSet<Integer>());
                }
            }
            int max = GCD(grid[0][0], grid[n - 1][m - 1]);
            int ans = 1;
            for(int i = 1; i * i <= max; i++){
                boolean[][] visited = new boolean[n][m];
                if(max % i != 0) continue;
                ArrayDeque<Pair> pq = new ArrayDeque<Pair>();
                pq.add(new Pair(0, 0));
                while(!pq.isEmpty()){
                    Pair cur = pq.poll();
                    if(visited[cur.x][cur.y]) continue;
                    else visited[cur.x][cur.y] = true;
                    if(cur.x == n - 1 && cur.y == m - 1){
                        ans = Math.max(ans, i);
                        break;
                    }
                    if(cur.x + 1 < n){
                        if(!visited[cur.x + 1][cur.y] && grid[cur.x + 1][cur.y] % i == 0) pq.add(new Pair(cur.x + 1, cur.y));
                    }
                    if(cur.y + 1 < m){
                        if(!visited[cur.x][cur.y + 1] && grid[cur.x][cur.y + 1] % i == 0) pq.add(new Pair(cur.x, cur.y + 1));
                    }
                }
                visited = new boolean[n][m];
                pq.clear();
                pq.add(new Pair(0, 0));
                while(!pq.isEmpty()){
                    Pair cur = pq.poll();
                    if(visited[cur.x][cur.y]) continue;
                    else visited[cur.x][cur.y] = true;
                    if(cur.x == n - 1 && cur.y == m - 1){
                        ans = Math.max(ans, max/i);
                        break;
                    }
                    if(cur.x + 1 < n){
                        if(!visited[cur.x + 1][cur.y] && grid[cur.x + 1][cur.y] % (max / i) == 0) pq.add(new Pair(cur.x + 1, cur.y));
                    }
                    if(cur.y + 1 < m){
                        if(!visited[cur.x][cur.y + 1] && grid[cur.x][cur.y + 1] % (max / i) == 0) pq.add(new Pair(cur.x, cur.y + 1));
                    }
                }
            }

            pw.println(ans);
        }
        pw.close();
    }

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

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

**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 the following points:

1. **Edge Cases**: Test cases that push the limits of the algorithm, such as grids with all the same values, grids with prime numbers, or grids with very large numbers.
2. **Path Constraints**: Ensure that the path from the top-left to the bottom-right corner is not straightforward, and there are multiple paths to consider.
3. **GCD Calculation**: The test case should ensure that the GCD calculation is not trivial and requires careful consideration of the path.

Given these points, let's create a test input that includes a grid where the GCD calculation is non-trivial and the path constraints are challenging.

```plaintext
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
```

This test input includes three test cases:

1. **2x3 Grid**:
   - Path from (0,0) to (1,2) with values 30, 20, 30, 15, 25, 40.
   - The maximum GCD is 10.

2. **3x3 Grid**:
   - Path from (0,0) to (2,2) with values 12, 4, 9, 3, 12, 2, 8, 3, 12.
   - The maximum GCD is 3.

3. **2x4 Grid**:
   - Path from (0,0) to (1,3) with values 2, 4, 6, 8, 1, 3, 6, 9.
   - The maximum GCD is 1.

This test input is designed to challenge the algorithm by including different grid sizes and values that require careful GCD calculations and path evaluations. The expected outputs are provided in the comments for verification.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

Buggy Code:
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define clr(x,val) memset((x),(val),sizeof(x))
ll gcd(ll a, ll b) { return ((b == 0) ? a : gcd(b, a % b)); }
ll lcm(ll a, ll b) { return (b / gcd(a, b)) * a; }
ll Ceil(ll a, ll b) { return (a + (b - 1)) / b; }
void hhh() {//freopen("input.txt", "r", stdin);//freopen("output.txt", "w", stdout);
    ios_base::sync_with_stdio(false); cin.tie(nullptr), cout.tie(nullptr);
}
int dx[] = { 0,0,-1,1,1,1,-1,-1 };
int dy[] = { 1,-1,0,0,-1,1,1,-1 };
double dist(int x1, int y1, int x2, int y2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }
bool getBit(ll num, int idx) { return ((num >> idx) & 1); }
ll setBit(ll num, int idx, bool val) { return val ? (num | (1LL << idx)) : (num & ~(1LL << idx)); }
ll flipBit(ll num, int idx) { return (num ^ (1LL << idx)); }
ll count_ones(ll n) { ll cnt = 0; while (n) { cnt++, n &= n - 1; }return cnt; }
int n, m,test=1;
vector<vector<int>>v;
map<int,int> dp[109][109];
map<int,int> vis[109][109];
bool valid(int i, int j) {
    return i >= 0 && i < n&& j >= 0 && j < m;
}
int solve(int i, int j, int g) {
    if (i == n - 1 && j == m - 1) {
        return g;
    }
    if (g == 1)return 1;
    int& ret = dp[i][j][g];
    if (vis[i][j][g] == test)return ret;
    vis[i][j][g] = test;
    ret = 1;
    if (valid(i + 1, j)) {
        ret = max(ret, solve(i + 1, j, gcd(g, v[i + 1][j])));
    }
    if (valid(i, j + 1)) {
        ret = max(ret, solve(i, j + 1, gcd(g, v[i][j + 1])));
    }
    return ret;
}
int main() {
    hhh();
    int T; cin >> T;
    while (T--) {
        cin >> n >> m;
        v = vector<vector<int>>(n, vector<int>(m));
        for (auto& it : v) {
            for (auto& itt : it)cin >> itt;
        }
        cout << solve(0, 0, v[0][0]) << "\n";
        test++;
    }
    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 given code, we need to consider edge cases and scenarios that might cause the algorithm to fail. The code uses a dynamic programming approach to find the maximum GCD along a path from the top-left to the bottom-right corner of the grid. The main issue in the code is the handling of the GCD values and the memoization.

### Step-by-Step Reasoning:

1. **Edge Cases:**
   - **Single Cell Grid:** A grid with only one cell.
   - **Straight Path:** A grid where the path from the top-left to the bottom-right is a straight line (either all down or all right).
   - **High GCD Values:** A grid where the GCD of the path can be very high, potentially causing issues with memoization.

2. **High GCD Values:**
   - If the GCD of the path is very high, the memoization might not work as expected, especially if the GCD values are not handled correctly.

3. **Memoization Issues:**
   - The memoization table `dp` and `vis` might not be reset properly between test cases, leading to incorrect results.

### Challenging Test Input:

Let's create a small grid where the GCD of the path is high and the path is straightforward. This will help in identifying if the memoization and GCD handling are correct.

```plaintext
3
1 1
10
2 2
10 20
30 40
3 3
10 20 30
40 50 60
70 80 90
```

### Explanation:

1. **Test Case 1:**
   - A 1x1 grid with a single cell containing 10.
   - The GCD is 10.

2. **Test Case 2:**
   - A 2x2 grid with the following values:
     - (0,0): 10
     - (0,1): 20
     - (1,0): 30
     - (1,1): 40
   - The GCD of the path (10 -> 20 -> 40 or 10 -> 30 -> 40) is 10.

3. **Test Case 3:**
   - A 3x3 grid with the following values:
     - (0,0): 10
     - (0,1): 20
     - (0,2): 30
     - (1,0): 40
     - (1,1): 50
     - (1,2): 60
     - (2,0): 70
     - (2,1): 80
     - (2,2): 90
   - The GCD of the path (10 -> 20 -> 30 -> 60 -> 90 or 10 -> 40 -> 70 -> 90) is 10.

This test input should help in exposing any issues with the GCD calculation and memoization in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

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

typedef long long ll;

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

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
        {
            if (i == 1 && j == 1)
                dp[i][j].insert(a[i][j]);
            if (j >= 1)
                for (int x : dp[i][j - 1])
                    dp[i][j].insert(gcd(x, a[i][j]));
            if (i >= 1)
                for (int x : dp[i - 1][j])
                    dp[i][j].insert(gcd(x, a[i][j]));
        }

    cout << *dp[n][m].rbegin() << '\n';
}

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

    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

Buggy Code:
import sys,math,bisect,heapq
from collections import deque
input=sys.stdin.readline
flush=sys.stdout.flush
mod=(10**9)+7
def merge(a,b,x):
    c=[]
    for i in a:
        c.append(math.gcd(x,i))
    for i in b:
        c.append(math.gcd(x,i))
    c.sort()
    t=[]
    for i in c:
        if t:
            if t[-1]!=i:
                t.append(i)
        else:
            t.append(i)
    return t
for _ in range(int(input())):
    #n=int(input())
    n,m=map(int,input().split())
    dp=[[]]*(m+1)
    dp[1]=[0]
    for i in range(n):
        a=list(map(int,input().split()))
        t=[[]]
        for j in range(m):
            t.append(merge(dp[j+1],t[j],a[j]))
        dp=t
    print(dp[-1][-1])

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases that can cause the algorithm to fail. The key aspects to consider are:

1. **Grid Size**: Small grids are easier to manage and validate.
2. **Grid Values**: Values that can lead to different GCD results depending on the path taken.
3. **Path Constraints**: Ensure that the paths from the top-left to the bottom-right corner can lead to different GCD values.

Let's consider a 3x3 grid with values that can create different GCD results based on the path taken. We will use values that are multiples of small prime numbers to make the GCD calculation interesting.

### Test Input
```plaintext
1
3 3
6 15 10
12 20 25
18 30 5
```

### Explanation
- **Grid Size**: 3x3, which is small enough to manage and validate.
- **Grid Values**:
  - The first row: 6, 15, 10
  - The second row: 12, 20, 25
  - The third row: 18, 30, 5

### Paths and GCDs
- **Path 1**: (0,0) -> (1,0) -> (2,0) -> (2,1) -> (2,2)
  - GCD: gcd(6, 12, 18, 30, 5) = 1
- **Path 2**: (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2)
  - GCD: gcd(6, 15, 10, 25, 5) = 5
- **Path 3**: (0,0) -> (0,1) -> (1,1) -> (2,1) -> (2,2)
  - GCD: gcd(6, 15, 20, 30, 5) = 5

### Expected Output
The maximum possible GCD along any path from the top-left to the bottom-right corner is 5.

This test input should help expose any issues in the provided buggy code, especially in how it handles the merging and GCD calculations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

Buggy Code:
import sys

# def gcd(x, y):
#     while y:
#         x, y = y, x % y
#     return x

def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)

def comding():
    n, m = map(int, input().split())
    a = [list(map(int, input().split())) for _ in range(n)]
    dp = [[set() for _ in range(m)] for _ in range(n)]
    dp[0][0].add(a[0][0])

    for i in range(n):
        for j in range(m):
            if i > 0:
                dp[i][j].update(gcd(x, a[i][j]) for x in dp[i - 1][j])
            if j > 0:
                dp[i][j].update(gcd(x, a[i][j]) for x in dp[i][j - 1])

    print(max(dp[n - 1][m - 1]))

if __name__ == "__main__":
    input = sys.stdin.readline
    for _ in range(int(input())):
        comding()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that can highlight issues in the dynamic programming approach used. Here are some considerations:

1. **Single Element Grid**: A grid with only one element (1x1) should always return that element as the GCD.
2. **Grid with All Elements Being the Same**: This should return the element itself as the GCD.
3. **Grid with Elements Having a Common Factor**: This can test the GCD calculation.
4. **Grid with Elements Having No Common Factor**: This should return 1 as the GCD.
5. **Grid with Large Values**: This can test the limits of the GCD function and the dynamic programming approach.

Let's create a test input that includes a combination of these scenarios to maximize the failure probability:

```plaintext
4
1 1
10
2 2
10 20
30 40
3 3
10 20 30
15 25 40
20 30 50
2 3
1 2 3
4 5 6
```

Explanation:
1. **1x1 Grid**: The GCD should be the single element itself, which is 10.
2. **2x2 Grid with Common Factors**: The GCD should be 10.
3. **3x3 Grid with Common Factors**: The GCD should be 5.
4. **2x3 Grid with No Common Factors**: The GCD should be 1.

This input will help in identifying any issues with the GCD calculation and the dynamic programming approach used 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: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

Buggy Code:
import java.util.*;

public class on_a_grid {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt();
            int m=sc.nextInt();
            int [][]arr=new int[n][m];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    arr[i][j]=sc.nextInt();
                }
            }
            Set<Integer>[][]g=new TreeSet[n+1][m+1];
            for (int i = 0; i <= n; i++) {
                Arrays.setAll(g[i], j -> new TreeSet<>());
            }
            g[1][1].add(gcd(arr[0][0],arr[n-1][m-1]));
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <=m ; j++) {
                    for(int x:g[i-1][j]){
                        g[i][j].add(gcd(x,arr[i-1][j-1]));
                    }
                    for(int x:g[i][j-1]){
                        g[i][j].add(gcd(x,arr[i-1][j-1]));
                    }
                    //System.out.println(g[i][j]);
                }
            }
            int ans=0;
            for (int x:g[n][m]) {
                ans=Math.max(x,ans);
            }
            System.out.println(ans);

        }
    }
    public static int gcd(int a,int b){
        return b==0?a:gcd(b,a%b);
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: GCD on a grid
time_limit_ms: 3000
memory_limit_mb: 256
Description: Not long ago, Egor learned about the Euclidean algorithm for finding the greatest common divisor of two numbers. The greatest common divisor of two numbers $$$a$$$ and $$$b$$$ is the largest number that divides both $$$a$$$ and $$$b$$$ without leaving a remainder. With this knowledge, Egor can solve a problem that he once couldn't.

Vasily has a grid with $$$n$$$ rows and $$$m$$$ columns, and the integer $$${a_i}_j$$$ is located at the intersection of the $$$i$$$-th row and the $$$j$$$-th column. Egor wants to go from the top left corner (at the intersection of the first row and the first column) to the bottom right corner (at the intersection of the last row and the last column) and find the greatest common divisor of all the numbers along the path. He is only allowed to move down and to the right. Egor has written down several paths and obtained different GCD values. He became interested in finding the maximum possible GCD.

Unfortunately, Egor is tired of calculating GCDs, so he asks for your help in finding the maximum GCD of the integers along the path from the top left corner to the bottom right corner of the grid.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 100$$$) — the number of rows and columns of the grid.

Then, there are $$$n$$$ lines, where the $$$i$$$-th line contains $$$m$$$ integers $$$(1 \le a_{i,j} \le {10}^{6}$$$) — the integers written in the $$$i$$$-th row and the $$$j$$$-th column of the grid.

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

Output: For each test case, output the maximum possible GCD along the path from the top left cell to the bottom right cell in a separate line.

Examples:
input:
3
2 3
30 20 30
15 25 40
3 3
12 4 9
3 12 2
8 3 12
2 4
2 4 6 8
1 3 6 9
output:
10
3
1

Note:

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

public class Main {
    static RealFastReader sc = new RealFastReader(System.in);
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        //int n=sc.ni();
        //double d=sc.nd();
        //String s=sc.ns();
        int T=sc.ni();
        while(T-->0){
            solve();
        }

        out.close();
    }

    public static void solve() {
        int n=sc.ni();
        int m=sc.ni();
        int arr[][]=new int[n][m];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++) arr[i][j]=sc.ni();
        }
        if(gcd(arr[0][0],arr[n-1][m-1])==1){
            System.out.println(1);
            return;
        }
        Set<Integer>[][] dp=new Set[n][m];
        for(int i=0;i<n;i++) {
            for (int j = 0; j < m; j++) {
                dp[i][j]=new HashSet<>();
            }
        }
        dp[0][0].add(arr[0][0]);

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(i>0){

                    for(int in:dp[i-1][j]){
                        int x=gcd(in,arr[i][j]);
                        dp[i][j].add(x);
                    }

                }
                if(j!=0){
                    for(int in:dp[i][j-1]){
                        int x=gcd(in,arr[i][j]);
                        dp[i][j].add(x);
                    }
                }

            }
        }
        int res=0;
        for(int in :dp[n-1][m-1]){
            res=Math.max(res,in);
        }
        System.out.println(res);

    }

public static int gcd(int a,int b){
        return b==0?a:gcd(b,a%b);
}
    //快读快写---------------------------------------------------------------------------------
    public static class RealFastReader {
        InputStream is;

        public RealFastReader(final InputStream is) {
            this.is = is;
        }

        private byte[] inbuf = new byte[8192];
        public int lenbuf = 0, ptrbuf = 0;

        public int readByte() {
            if (lenbuf == -1) {
                throw new InputMismatchException();
            }
            if (ptrbuf >= lenbuf) {
                ptrbuf = 0;
                try {
                    lenbuf = is.read(inbuf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (lenbuf <= 0) {
                    return -1;
                }
            }
            return inbuf[ptrbuf++];
        }

        private boolean isSpaceChar(int c) {
            return !(c >= 33 && c <= 126);
        }

        private int skip() {
            int b;
            while ((b = readByte()) != -1 && isSpaceChar(b)) ;
            return b;
        }

        public double nd() {
            return Double.parseDouble(ns());
        }

        public char nc() {
            return (char) skip();
        }

        public String ns() {
            int b = skip();
            StringBuilder sb = new StringBuilder();
            while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }

        public char[] ns(int n) {
            char[] buf = new char[n];
            int b = skip(), p = 0;
            while (p < n && !(isSpaceChar(b))) {
                buf[p++] = (char) b;
                b = readByte();
            }
            return n == p ? buf : Arrays.copyOf(buf, p);
        }

        public int[] na(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public long[] nal(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        }

        public char[][] nm(int n, int m) {
            char[][] map = new char[n][];
            for (int i = 0; i < n; i++) {
                map[i] = ns(m);
            }
            return map;
        }

        public int[][] nmi(int n, int m) {
            int[][] map = new int[n][];
            for (int i = 0; i < n; i++) {
                map[i] = na(m);
            }
            return map;
        }

        public int ni() {
            int num = 0;
            int b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }

        public long nl() {
            long num = 0;
            int b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to create a test case that highlights the issue with the algorithm. The bug in the code is that it prematurely checks if the GCD of the top-left and bottom-right corners is 1 and returns 1 if that is the case. This approach is incorrect because there might be other paths that yield a higher GCD value.

### Step-by-Step Reasoning:
1. **Identify the Bug**: The bug is in the early return statement:
   ```java
   if(gcd(arr[0][0],arr[n-1][m-1])==1){
       System.out.println(1);
       return;
   }
   ```
   This statement assumes that if the GCD of the top-left and bottom-right corners is 1, the maximum GCD for any path must be 1. However, this is not always true.

2. **Create a Counterexample**: We need a grid where the GCD of the top-left and bottom-right corners is 1, but there exists a path with a GCD greater than 1.

3. **Construct the Grid**:
   - Let's create a 2x2 grid where the top-left and bottom-right corners are 1, but the other cells are such that a path exists with a GCD greater than 1.

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

### Explanation:
- **Grid**:
  ```
  1 2
  3 1
  ```
- **Paths**:
  - Path 1: (1, 1) -> (1, 2) -> (2, 2) -> GCD = gcd(1, 2, 1) = 1
  - Path 2: (1, 1) -> (2, 1) -> (2, 2) -> GCD = gcd(1, 3, 1) = 1

- **Expected Output**:
  - The maximum GCD for any path is 1, but the bug in the code will incorrectly return 1 without considering other paths.

This test case should expose the bug in the provided code. The code will return 1, but it should not make the early return and should instead compute the GCD for all possible paths.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

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

int main(){
    int t;
    cin>>t;
    while(t--){
        int a,b,c;
        cin>>a>>b>>c;
        if(a>b && a>c) {
            int p=5;
            while(p--){
                if(b>=c) c++;
                else b++;
            }
            cout<<a*b*c<<endl;
        }
        else if(b>a && b>c) {
            int p=5;
            while(p--){
                if(a>=c) c++;
                else a++;
            }
            cout<<a*b*c<<endl;
        }
        else{
            int p=5;
            while(p--){
                if(b>=a) a++;
                else b++;
            }
            cout<<a*b*c<<endl;
        }
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

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

#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;

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

void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(lld t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(ull t) {cerr << t;}

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.first); cerr << ","; _print(p.second); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

void solve() {
   int a,b,c;
   cin>>a>>b>>c;
   int avg = a+b+c+5;
   int req = avg/3;
   if(avg%3 == 0){
    req++;
   }
   int k = 5;
   vector<int>v;
   v.push_back(a);
   v.push_back(b);
   v.push_back(c);
   sort(v.begin(),v.end());
   while(k--){
    if(v[0]<=req){
        v[0]+=1;
    }else if(v[1] <= req){
        v[1]+=1;
    }else{
        v[2]+=1;
    }
   }
   cout<<v[0]*v[1]*v[2]<<endl;
}

int main() {
    fastio();
#ifndef ONLINE_JUDGE
    freopen("Error.txt", "w", stderr);
#endif
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

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

using namespace std;

int main() {
    int n;
    cin >> n;
    for (int q = 0; q < n; q++) {
        vector <int> d(3);
        for (int i = 0; i < 3; i++) {
            cin >> d[i];
        }
        sort(d.begin(), d.end());
        int c = 0, k = 0;
        while (k < 5) {
            for (int i = 0; i < 3; i++) {
                if (i != 2) {
                    while (d[i] < d[i+1]) {
                        d[i]++;
                        k++;
                        if (k > 4) {
                            break;
                        }
                    }
                }
                if (k > 4) {
                    break;
                }
            }
            //cout << d[0] << " " << d[1] << " " << d[2] << " " << k << " ";
            if (k < 5 && d[0] == d[1] && d[0] == d[2]) {
                for (int i = 0; i < 5-k; i++) {
                    d[(i+3)%3]++;
                }
                break;
            }
        }
        cout << d[0] * d[1] * d[2] << endl;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

Buggy Code:
#!/usr/bin/python3

if __name__ == '__main__':
    t = 1
    t = int(input())

    for _ in range(t):
        a, b, c = map(int, input().split())
        # get max of a * b * c and we have 5 ones to add
        arr = sorted([a, b, c])
        cnt = 5

        while cnt > 0 and arr[0] < arr[1]:
            arr[0] += 1
            cnt -= 1
        while cnt > 1 and arr[1] < arr[2]:
            arr[0] += 1
            arr[1] += 1
            cnt -= 2
        if cnt == 1:
            arr[2] += 1
        while cnt > 0:
            arr[0] += 1
            cnt -= 1
            if cnt == 0:
                break
            arr[1] += 1
            cnt -= 1
            if cnt == 0:
                break
            arr[2] += 1
            cnt -= 1
        print(arr[0] * arr[1] * arr[2])

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

Buggy Code:
t = int(input())
for i in range(t):
    l = list(map(int, input().split()))
    l.sort()
    if l[0]+5 - l[2] < 0:
        print((l[0]+5)*l[1]*l[2])
    else:
        print((l[2]+1) * (l[1] + 5 - (l[2]+1 - l[0])) * l[2])

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

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

Think step by step.

Response:
```plaintext
3
1 1 1
9 1 1
5 5 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

Buggy Code:
t=int(input())
for i in range(t):
    a_b_c0=input()
    a_b_clist=a_b_c0.split()
    a_b_c=[int(x) for x in a_b_clist]
    #print(a_b_c)
    count=0
    #print(maxx)
    while(count<5):

        for j in range(3):
            minn=min(a_b_c)
            if(a_b_c[j]==minn):
                a_b_c[j]+=1
                count+=1
    #print(a_b_c)
    output=a_b_c[0]*a_b_c[1]*a_b_c[2]
    print(output)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

Buggy Code:
import java.util.Scanner;
import java.util.Arrays;

public class MaxBananas {

    public static int maxProduct(int[] abc) {
        Arrays.sort(abc);
        int increasesLeft = 5;

        if (increasesLeft > 0) {
            int increase = Math.min(increasesLeft, abc[1] - abc[0]);
            abc[0] += increase;
            increasesLeft -= increase;
        }

        if (increasesLeft > 0) {
            int increase = Math.min(increasesLeft, abc[2] - abc[1]);
            abc[1] += increase;
            increasesLeft -= increase;
        }

        if (increasesLeft > 0) {
            abc[0] += increasesLeft / 3;
            abc[1] += increasesLeft / 3;
            abc[2] += increasesLeft / 3;
            int remainder = increasesLeft % 3;

            if (remainder == 1) {
                abc[0] += 1;
            } else if (remainder == 2) {
                abc[0] += 1;
                abc[1] += 1;
            }
        }

        return abc[0] * abc[1] * abc[2];
    }

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

        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            int a = scanner.nextInt();
            int b = scanner.nextInt();
            int c = scanner.nextInt();

            int[] abc = {a, b, c};
            int result = maxProduct(abc);
            System.out.println(result);
        }

        scanner.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Only Pluses
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kmes has written three integers $a$, $b$ and $c$ in order to remember that he has to give Noobish_Monk $a \times b \times c$ bananas.

Noobish_Monk has found these integers and decided to do the following at most $5$ times:

  * pick one of these integers;    * increase it by $1$.

For example, if $a = 2$, $b = 3$ and $c = 4$, then one can increase $a$ three times by one and increase $b$ two times. After that $a = 5$, $b = 5$, $c = 4$. Then the total number of bananas will be $5 \times 5 \times 4 = 100$.

What is the maximum value of $a \times b \times c$ Noobish_Monk can achieve with these operations?

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of the test cases follows.

The first and only line of each test case contains three integers $a$, $b$ and $c$ ($1 \le a, b, c \le 10$) — Kmes's integers.

Output: For each test case, output a single integer — the maximum amount of bananas Noobish_Monk can get.

Examples:
input:
2
2 3 4
10 1 10
output:
100
600

Note:

Buggy Code:
import java.util.*;

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

        // taking no. of testcases as input
        int t = sc.nextInt();

        while (t-- > 0) {
            // taking a, b, c as input
            int a = sc.nextInt();
            int b = sc.nextInt();
            int c = sc.nextInt();

            // finding max and min
            int max = Math.max(Math.max(a, b), c);
            int min = Math.min(Math.min(a, b), c);

            int constant = 5;

            while (min != max && constant > 0) {
                if (a < max && constant > 0) {
                    a++;
                    constant--;
                }
                if (b < max && constant > 0) {
                    b++;
                    constant--;
                }
                if (c < max && constant > 0) {
                    c++;
                    constant--;
                }
                max = Math.max(Math.max(a, b), c);
                min = Math.min(Math.min(a, b), c);
            }
            switch (constant) {
                case 1:
                    a++;
                    constant = 0;
                    break;
                case 2:
                    a++;
                    b++;
                    constant = 0;
                    break;
                case 3:
                    a++;
                    b++;
                    c++;
                    constant = 0;
                    break;
                case 4:
                    a += 2;
                    b++;
                    c++;
                    constant = 0;
                    break;
                case 5:
                    a += 2;
                    b += 2;
                    c++;
                    constant = 0;
                    break;
                default:
                    break;
            }

            System.out.println(a * b * c);
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 1 1
9 9 9
2 9 9
```