#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
using LL = long long;
LL a[305][305];
vector<pair<int, int> > g[605];
LL d[605];

int main(){

#ifdef LOCAL
    freopen("data.in", "r", stdin);
    freopen("data.out", "w", stdout);
#endif

    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

    int T;
    cin >> T;
    while(T--){
        int n, m;
        cin >> n >> m;
        for(int i = 1; i <= n - 1; i++){
            for(int j = 1; j <= m - 1; j++){
                cin >> a[i][j];
            }
        }
        for(int i = n - 1; i >= 1; i--){
            for(int j = m - 1; j >= 1; j--){
                a[i][j] -= a[i + 1][j] + a[i][j + 1] + a[i + 1][j + 1];
            }
        }
        // for(int i = 1; i <= n; i++){
        //     for(int j = 1; j <= m; j++){
        //         cout << a[i][j] << " \n"[j == m];
        //     }
        // }
        for(int i = 1; i <= n + m; i++) g[i].clear();
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                // 0 <= row(i) - col(j) + a[i][j] <= 1000'000
                if ((i + j) % 2 == 0){
                    g[j + n].push_back({i, 1000'00 - a[i][j]});
                    g[i].push_back({j + n, a[i][j]});
                }
                // 0 <= col(j) - row(i) + a[i][j] <= 1000'000
                else{
                    g[i].push_back({j + n, 1000'00 - a[i][j]});
                    g[j + n].push_back({i, a[i][j]});
                }
            }
        }
        memset(d, 0x3f, sizeof d);
        d[1] = 0;
        for(int _ = 0; _ < n + m - 1; _++){
            for(int i = 1; i <= n + m; i++){
                for(auto [j, w] : g[i]){
                    d[j] = min(d[j], d[i] + w);
                }
            }
        }
        bool ok = true;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                if ((i + j) % 2 == 0) a[i][j] += d[i];
                else a[i][j] -= d[i];
                if ((i + j) % 2 == 0) a[i][j] -= d[j + n];
                else a[i][j] += d[j + n];
                ok &= (a[i][j] >= 0) and (a[i][j] <= 1000'000);
            }
        }
        if (ok){
            cout << "YES" << '\n';
            for(int i = 1; i <= n; i++){
                for(int j = 1; j <= m; j++){
                    cout << a[i][j] << " \n"[j == m];
                }
            }
        }
        else{
            cout << "NO" << '\n';
        }
    }

}