#include <bits/stdc++.h>

using namespace std;

ifstream f("plantatie.in");
ofstream g("plantatie.out");

int n,m,i,j,x,k,pas,p,maxim,A[510][510][15];

int main()
{
    // RMQ pe matrice
    // A[i][j][k] = maximul valorilor determinate de patratul ce are coltul stanga-sus in (i;j) si lungimea laturii = 2^k
            // Adica patratul (i;j) pana in (i + 2^k -1; j + 2^k -1);

    f>>n>>m;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
        {
            f>>x;
            A[i][j][0]=x;
        }
    }

    // Construim A
    // A[i][j][k] = maximul dintre patratul [ (i;j) -> (i + 2^(k-1) - 1; j + 2^(k-1) - 1) ]
    //                                      [ ( i;j + 2^(k-1) ) -> (i + 2^(k-1) - 1; j + 2^(k-1) + 2^(k-1) - 1)] = [ ( i;j + 2^(k-1) ) -> (i + 2^(k-1) - 1; j + 2^k - 1)]
    //                                      [ ( i + 2^(k-1);j ) -> (i + 2^(k-1) + 2^(k-1) - 1; j + 2^(k-1) - 1)] = [ ( i + 2^(k-1);j ) -> (i + 2^k - 1; j + 2^(k-1) - 1)]
    //                                      [ ( i + 2^(k-1);j + 2^(k-1) ) -> (i + 2^k - 1; j + 2^k - 1) ]

                                                        // TOATE ACESTE PATRATE SE POT INTERSECTA, DAR REUNIUNEA LOR DA PATRATUL DORIT

    for(k=1;(1<<k)<=n;k++)
    {
        for(i=1;i+(1<<k)-1<=n;i++)
        {
            for(j=1;j+(1<<k)-1<=n;j++)
            {
                maxim=max(A[i][j][k-1], A[i][j+(1<<(k-1))][k-1]);
                maxim=max(maxim, A[i+(1<<(k-1))][j][k-1]);
                maxim=max(maxim, A[i+(1<<(k-1))][j+(1<<(k-1))][k-1]);

                A[i][j][k]=maxim;
            }
        }
    }

    // REZOLVAM
    for(pas=1;pas<=m;pas++)
    {
        f>>i>>j>>k;

        // cautam un p, astfel incat 2^p <= lungime = k
        // iar solutia o sa fie reuniunea patratelor [ (i;j) -> (i + 2^p - 1; j + 2^p - 1) ] --- stanga sus
        //                                           [ (i;j + k - 2^p) -> (i + 2^p - 1; j + k - 2^p + 2^p - 1) ] = [ (i;j - 2^p + 1) -> (i + 2^p -1; j + k -1) ] --- dreapta sus
        //                                           [ (i + k - 2^p;j) -> (i + k - 1; j + 2^p - 1) ] --- stanga jos
        //                                           [ (i + k - 2^p;j + k - 2^p) -> (i+k-1; j+k-1)] --- dreapta jos

        p=log2(k);

        maxim=max(A[i][j][p], A[i][j+k-(1<<p)][p]);
        maxim=max(maxim, A[i+k-(1<<p)][j][p]);
        maxim=max(maxim, A[i+k-(1<<p)][j+k-(1<<p)][p]);

        g<<maxim<<'\n';
    }
    return 0;
}