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 int N = (int)1e6 + 5;
const int mod = 1000000007;
int mat[11][101][101];
int main() {
  int n, q, c, i, j, k;
  scanf("%d", &n);
  scanf("%d", &q);
  scanf("%d", &c);
  c++;
  for (i = 1; i <= n; i++) {
    int x, y, s;
    scanf("%d", &x);
    scanf("%d", &y);
    scanf("%d", &s);
    mat[s][x][y]++;
  }
  int x, y;
  for (x = 0; x <= c - 1; x++) {
    for (i = 1; i <= 100; i++) {
      for (j = 1; j <= 100; j++) {
        mat[x][i][j] +=
            (mat[x][i][j - 1] + mat[x][i - 1][j] - mat[x][i - 1][j - 1]);
      }
    }
  }
  int t, x2, y2;
  while (q--) {
    scanf("%d", &t);
    scanf("%d", &x);
    scanf("%d", &y);
    scanf("%d", &x2);
    scanf("%d", &y2);
    long long ans = 0;
    for (i = 0; i <= c - 1; i++) {
      ans += ((i + t) % c) * (mat[i][x2][y2] + mat[i][x - 1][y - 1] -
                              mat[i][x - 1][y2] - mat[i][x2][y - 1]);
    }
    printf("%lld ", ans);
    printf("\n");
  }
  return 0;
}
