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#include <fstream>
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
using namespace std;
class InParser {
private:
FILE *fin;
char *buff;
int sp;
char read_ch() {
++sp;
if (sp == 4096) {
sp = 0;
fread(buff, 1, 4096, fin);
}
return buff[sp];
}
public:
InParser(const char* nume) {
fin = fopen(nume, "r");
buff = new char[4096]();
sp = 4095;
}
InParser& operator >> (int &n) {
char c;
while (!isdigit(c = read_ch()) && c != '-');
int sgn = 1;
if (c == '-') {
n = 0;
sgn = -1;
} else {
n = c - '0';
}
while (isdigit(c = read_ch())) {
n = 10 * n + c - '0';
}
n *= sgn;
return *this;
}
InParser& operator >> (long long &n) {
char c;
n = 0;
while (!isdigit(c = read_ch()) && c != '-');
long long sgn = 1;
if (c == '-') {
n = 0;
sgn = -1;
} else {
n = c - '0';
}
while (isdigit(c = read_ch())) {
n = 10 * n + c - '0';
}
n *= sgn;
return *this;
}
} f("plantatie.in");
class OutParser {
private:
FILE *fout;
char *buff;
int sp;
void write_ch(char ch) {
if (sp == 50000) {
fwrite(buff, 1, 50000, fout);
sp = 0;
buff[sp++] = ch;
} else {
buff[sp++] = ch;
}
}
public:
OutParser(const char* name) {
fout = fopen(name, "w");
buff = new char[50000]();
sp = 0;
}
~OutParser() {
fwrite(buff, 1, sp, fout);
fclose(fout);
}
OutParser& operator << (int vu32) {
if (vu32 <= 9) {
write_ch(vu32 + '0');
} else {
(*this) << (vu32 / 10);
write_ch(vu32 % 10 + '0');
}
return *this;
}
OutParser& operator << (long long vu64) {
if (vu64 <= 9) {
write_ch(vu64 + '0');
} else {
(*this) << (vu64 / 10);
write_ch(vu64 % 10 + '0');
}
return *this;
}
OutParser& operator << (char ch) {
write_ch(ch);
return *this;
}
OutParser& operator << (const char *ch) {
while (*ch) {
write_ch(*ch);
++ch;
}
return *this;
}
} g("plantatie.out");
int n, m, x, y, k, lg[505], a[10][505][505];
void RMQ()
{
for (int i = 1; i <= n; ++i) {
lg[i] = lg[i >> 1] + 1;
}
for (int k = 1; k <= lg[n]; ++k) {
int dr = (1 << (k - 1));
for (int i = 1; i <= n - dr; ++i) {
for (int j = 1; j <= n - dr; ++j) {
a[k][i][j] = max ( max(a[k - 1][i][j], a[k - 1][i + dr][j]), max (a[k - 1][i][j + dr], a[k - 1][i + dr][j + dr]) );
}
}
}
}
int main()
{
f >> n >> m;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
f >> a[0][i][j];
}
}
RMQ();
while (m--) {
f >> x >> y >> k;
int dr = (1 << (lg[k] - 1)), kk = lg[k] - 1;
g << max( max(a[kk][x][y], a[kk][x + k - dr][y]), max(a[kk][x][y + k - dr], a[kk][x + k - dr][y + k - dr]) ) << '\n';
}
return 0;
}