int main(){

    for (int i = ; i <= n; i++)

        for (int j = ; j <= m; j++) {

            scanf("%d", &val[i][j]);

            dp[i][j][][] = val[i][j];

        }

    for (int i = ; ( << i) <= n; i++) {

        for (int j = ; ( << j) <= m; j++) {

            if (i ==  && j == ) continue;

            for (int row = ; row + ( << i) -  <= n; row++)

                for (int col = ; col + ( << j) -  <= m; col++) {

                    //当x或y等于0的时候，就相当于一维的RMQ了

                    //if(i == 0) dp[row][col][i][j] = max(dp[row][col][i][j - 1], dp[row][col + (1 << (j - 1))][i][j - 1]);

                    if (j == ) dp[row][col][i][j] = max(dp[row][col][i - ][j], dp[row + ( << (i - ))][col][i - ][j]);

                    else dp[row][col][i][j] = max(dp[row][col][i][j - ], dp[row][col + ( << (j - ))][i][j - ]);

                }

        }

    }

}

//本来一维RMQ询问的时候是一个区间，现在变成了一个矩形，所以需要四个角度

int rmq(int x1, int y1, int x2, int y2) {

    int kx = , ky = ;

    while (( << ( + kx)) <= x2 - x1 + 1) kx++;

    while (( << ( + ky)) <= y2 - y1 + 1) ky++;

    int m1 = dp[x1][y1][kx][ky];

    int m2 = dp[x2 - ( << kx) + ][y1][kx][ky];

    int m3 = dp[x1][y2 - ( << ky) + ][kx][ky];

    int m4 = dp[x2 - ( << kx) + ][y2 - ( << ky) + ][kx][ky];

    return max(max(m1, m2), max(m3, m4));

}