权值排序，整体二分，没了。
// by Isaunoya
#include
using namespace std;
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
#define int long long
const int _ = 1 << 21;
struct I {
char fin[_], *p1 = fin, *p2 = fin;
inline char gc() {
return (p1 == p2) && (p2 = (p1 = fin) + fread(fin, 1, _, stdin), p1 == p2) ? EOF : *p1++;
}
inline I& operator>>(int& x) {
bool sign = 1;
char c = 0;
while (c < 48) ((c = gc()) == 45) && (sign = 0);
x = (c & 15);
while ((c = gc()) > 47) x = (x << 1) + (x << 3) + (c & 15);
x = sign ? x : -x;
return *this;
}
inline I& operator>>(double& x) {
bool sign = 1;
char c = 0;
while (c < 48) ((c = gc()) == 45) && (sign = 0);
x = (c - 48);
while ((c = gc()) > 47) x = x * 10 + (c - 48);
if (c == '.') {
double d = 1.0;
while ((c = gc()) > 47) d = d * 0.1, x = x + (d * (c - 48));
}
x = sign ? x : -x;
return *this;
}
inline I& operator>>(char& x) {
do
x = gc();
while (isspace(x));
return *this;
}
inline I& operator>>(string& s) {
s = "";
char c = gc();
while (isspace(c)) c = gc();
while (!isspace(c) && c != EOF) s += c, c = gc();
return *this;
}
} in;
struct O {
char st[100], fout[_];
signed stk = 0, top = 0;
inline void flush() {
fwrite(fout, 1, top, stdout), fflush(stdout), top = 0;
}
inline O& operator<<(int x) {
if (top > (1 << 20)) flush();
if (x < 0) fout[top++] = 45, x = -x;
do
st[++stk] = x % 10 ^ 48, x /= 10;
while (x);
while (stk) fout[top++] = st[stk--];
return *this;
}
inline O& operator<<(char x) {
fout[top++] = x;
return *this;
}
inline O& operator<<(string s) {
if (top > (1 << 20)) flush();
for (char x : s) fout[top++] = x;
return *this;
}
} out;
#define pb emplace_back
#define fir first
#define sec second
template < class T > inline void cmax(T & x , const T & y) {
(x < y) && (x = y) ;
}
template < class T > inline void cmin(T & x , const T & y) {
(x > y) && (x = y) ;
}
int n , m ;
const int N = 500 + 10 ;
const int M = 6e4 + 10 ;
struct point { int x , y , val ; } a[N * N] ;
struct que { int x , y , x2 , y2 , k , id ; } q[M] ;
int mcnt = 0 ;
int c[N][N] ;
int low(int x) { return x & -x ; }
void add(int x , int y , int v) {
for(int i = x ; i <= n ; i += low(i)) for(int j = y ; j <= n ; j += low(j)) c[i][j] += v ;
}
int qry(int x , int y) {
if(! x || ! y) return 0 ;
int ans = 0 ;
for(int i = x ; i > 0 ; i ^= low(i)) for(int j = y ; j > 0 ; j ^= low(j)) ans += c[i][j] ;
return ans ;
}
int qry(int x , int y , int x2 , int y2) {
return qry(x2 , y2) - qry(x - 1 , y2) - qry(x2 , y - 1) + qry(x - 1 , y - 1) ;
}
int ans[M] ;
que t1[M] , t2[M] ;
void solve(int l , int r , int ql , int qr) {
if(ql > qr) return ;
if(l == r) {
rep(i , ql , qr) ans[q[i].id] = a[l].val ;
return ;
}
int mid = l + r >> 1 ;
rep(i , l , mid) add(a[i].x , a[i].y , 1) ;
int cnt1 = 0 , cnt2 = 0 ;
rep(i , ql , qr) {
int res = qry(q[i].x , q[i].y , q[i].x2 , q[i].y2) ;
if(res >= q[i].k) t1[++ cnt1] = q[i] ;
else { t2[++ cnt2] = q[i] ; t2[cnt2].k -= res ; }
}
rep(i , l , mid) add(a[i].x , a[i].y , -1) ;
rep(i , 1 , cnt1) q[i + ql - 1] = t1[i] ;
rep(i , 1 , cnt2) q[ql + cnt1 + i - 1] = t2[i] ;
solve(l , mid , ql , ql + cnt1 - 1) ;
solve(mid + 1 , r , ql + cnt1 , qr) ;
}
signed main() {
#ifdef _WIN64
freopen("testdata.in" , "r" , stdin) ;
#endif
in >> n >> m ;
rep(i , 1 , n) rep(j , 1 , n) { int k ; in >> k ; a[++ mcnt] = { i , j , k } ; }
rep(i , 1 , m) { in >> q[i].x >> q[i].y >> q[i].x2 >> q[i].y2 >> q[i].k ; q[i].id = i ; }
sort(a + 1 , a + mcnt + 1 , [](point x , point y) { return x.val < y.val ; }) ;
solve(1 , mcnt , 1 , m) ; rep(i , 1 , m) out << ans[i] << '\n' ;
return out.flush(), 0;
}