#pragma GCC optimize(3,"Ofast","inline")
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
//typedef __int128 LL;//2的128
typedef long long ll; //2的64
// 1e9,, 1e19,,1e38
#define int long long
int n,m;
vector<vector<int>>a;
bool check(int x){
    vector<vector<int>>s(n+1,vector<int>(m+1,0));
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
    s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+(a[i][j]>=x);                   
        }//符合条件的矩阵都变成1
    }
    for(int i=x;i<=n;i++){
        for(int j=x;j<=m;j++){
        int z=i-x+1,y=j-x+1;
    int ti=s[i][j]-s[z-1][j]-s[i][y-1]+s[z-1][y-1];
    if(ti==x*x) return 1;   //如果这个区域全是>=x的那么就等于k*k
        }
    }
    return 0;
}
void solve()
{   cin>>n>>m;
    a=vector<vector<int>>(n+5,vector<int>(m+5,0));
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++ ) cin>>a[i][j];
    }
    int l=1,r=min(n,m);
        while(l<r){
        int mid=(l+r+1)/2;
        if(check(mid)) l=mid;
        else r=mid-1;
    }
    cout<<l<<endl;
}

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


    return 0;
}