Method 1:二分答案
------------------------------------------AC code------------------------------------------

#include<cstdio>

using namespace std;

const int MAXN = 2005;
int n, m, g[MAXN][MAXN];

inline int min(int a, int b){
    return a < b ? a : b;
}

inline int read(){
    int x = 0;
    bool fl = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9'){
        if(ch == '-')   fl = 0;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return fl ? x : -x;
}

bool check(int mid){
    for(int i = mid; i <= n; i++){
        for(int j = mid; j <= m; j++){
            int cnt = g[i][j] - g[i][j-mid] - g[i-mid][j] + g[i-mid][j-mid];
            if(cnt >= mid*mid)  return true;
        }
    }
    return false;
}

int main(){
//freopen("t1.in", "r", stdin);
//freopen("t1.out", "w", stdout);
    n = read(), m = read();
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            g[i][j] = read(), g[i][j] = g[i][j] + g[i-1][j] + g[i][j-1] - g[i-1][j-1];
    int l = 0, r = min(n, m)+1, ans = 0;
    while(l < r){
        int mid = l+(r-l)/2;
        if(check(mid))
            if(l == mid)    break;
            else    l = mid;
        else    r = mid;
    }
    printf("%d", l);
    return 0;
}