The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c).

Over time the stars twinkle. At moment 0 the i-th star has brightness si. Let at moment t some star has brightness x. Then at moment (t + 1) this star will have brightness x + 1, if x + 1 ≤ c, and 0, otherwise.

You want to look at the sky q times. In the i-th time you will look at the moment ti and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x1i, y1i) and the upper right — (x2i, y2i). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.

A star lies in a rectangle if it lies on its border or lies strictly inside it.

Input

The first line contains three integers n, q, c (1 ≤ n, q ≤ 105, 1 ≤ c ≤ 10) — the number of the stars, the number of the views and the maximum brightness of the stars.

The next n lines contain the stars description. The i-th from these lines contains three integers xi, yi, si (1 ≤ xi, yi ≤ 100, 0 ≤ si ≤ c ≤ 10) — the coordinates of i-th star and its initial brightness.

The next q lines contain the views description. The i-th from these lines contains five integers ti, x1i, y1i, x2i, y2i (0 ≤ ti ≤ 109, 1 ≤ x1i < x2i ≤ 100, 1 ≤ y1i < y2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.

Output

For each view print the total brightness of the viewed stars.

Examples

Input

2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5


Output

3
0
3


Input

3 4 5
1 1 2
2 3 0
3 3 1
0 1 1 100 100
1 2 2 4 4
2 2 1 4 7
1 50 50 51 51


Output

3
3
5
0

Note

Let's consider the first example.

At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.

At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.

At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long N = 2e5;
const long long inf = 1e18;
template <typename... T>
void read(T&... args) {
  ((cin >> args), ...);
}
template <typename... T>
void write(T&&... args) {
  ((cout << args << " "), ...);
}
template <typename... T>
void writeline(T&&... args) {
  ((cout << args << " "), ...);
  cout << "\n";
}
long long add(long long a, long long b) {
  long long res = (a + b) % mod;
  return (res < 0) ? res + mod : res;
}
long long mul(long long a, long long b) {
  long long res = (a * 1LL * b) % mod;
  return (res < 0) ? res + mod : res;
}
string binary(long long x) {
  string res = "";
  for (long long i = 0; i < 32; i++) res += (1 << i & x) ? '1' : '0';
  return res;
}
long long in(long long x, long long y, long long z) {
  if (y > z) swap(y, z);
  return (x >= y && x <= z);
}
struct customCompare {
  bool operator()(long long x, long long y) { return x > y; }
};
long long sky[105][105][11];
void fun() {
  for (long long i = 1; i < 105; i++) {
    for (long long j = 1; j < 105; j++) {
      for (long long k = 0; k < 11; k++) {
        sky[i][j][k] +=
            sky[i - 1][j][k] + sky[i][j - 1][k] - sky[i - 1][j - 1][k];
      }
    }
  }
}
void solve() {
  long long n, q, c;
  read(n, q, c);
  for (long long i = 0; i < n; i++) {
    long long x, y, s;
    read(x, y, s);
    sky[x][y][s]++;
  }
  fun();
  long long ans;
  for (long long i = 0; i < q; i++) {
    long long t, x1, y1, x2, y2;
    read(t, x1, y1, x2, y2);
    long long ans = 0;
    long long temp[c + 1];
    for (long long j = 0; j < c + 1; j++) {
      temp[j] = sky[x2][y2][j] - sky[x1 - 1][y2][j] - sky[x2][y1 - 1][j] +
                sky[x1 - 1][y1 - 1][j];
    }
    for (long long j = 0; j < c + 1; j++) {
      ans += temp[j] * ((j + t) % (c + 1));
    }
    writeline(ans);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
