Grid GCD, memset se TLE

/**
 *    author:  heWhoCooks
 *    created: 2024-12-27 19:50:38
 **/
#include <bits/stdc++.h>
using namespace std;
mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());
#define NeedForSpeed                  \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define int long long
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<vi> vvi;
typedef vector<pair<int, int>> vpi;
#define f first
#define s second
#define pb push_back
#define pf push_front
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define endl "\n"
#define M_PI 3.14159265358979323846L
const int mod = 1e9 + 7;
const int INF = 2e18;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
const int N = 105;
int dp[N][N], grid[N][N];
void solve()
{
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> grid[i][j];
        }
    }
    vi divisors;
    int maxi = gcd(grid[1][1], grid[n][m]); // have to be taken anyhow
    for (int i = 1; i * i <= maxi; i++)
    {
        if (maxi % i == 0)
        {
            divisors.pb(i);
            if (i != maxi / i)
            {
                divisors.pb(maxi / i);
            }
        }
    }
    sort(all(divisors), greater<int>());
    for (auto &div : divisors)
    {
        // reset, memset is giving TLE
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                dp[i][j] = 0;
            }
        }
        // check if we can reach the end with this gcd
        // base case, initial state true
        dp[1][1] = 1;
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                if (!dp[i][j])
                {
                    // can't reach here
                    continue;
                }
                if (i + 1 <= n && ((grid[i + 1][j] % div) == 0))
                {
                    dp[i + 1][j] = 1;
                }
                if (j + 1 <= m && ((grid[i][j + 1] % div) == 0))
                {
                    dp[i][j + 1] = 1;
                }
            }
        }
        if (dp[n][m])
        {
            cout << div << endl;
            return;
        }
    }
    cout << -1 << endl;
    return;
}

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