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>
const int maxn = 1000001;
using namespace std;
int main() {
  int n, q, c, t, i, j, k, x, y;
  int x1, x2, y1, y2, c1;
  int star[105][105][15];
  scanf("%d%d%d", &n, &q, &c);
  memset(star, 0, sizeof(star));
  for (i = 0; i < n; i++) {
    scanf("%d%d%d", &x, &y, &c1);
    star[x][y][c1]++;
  }
  for (i = 0; i <= 10; i++) {
    for (j = 1; j < 105; j++) {
      for (k = 1; k < 105; k++) {
        star[j][k][i] +=
            (star[j - 1][k][i] + star[j][k - 1][i] - star[j - 1][k - 1][i]);
      }
    }
  }
  while (q--) {
    scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2);
    int ans = 0;
    for (i = 0; i <= c; i++) {
      int cnt, light;
      light = (i + t) % (c + 1);
      cnt = star[x2][y2][i] + star[x1 - 1][y1 - 1][i] - star[x2][y1 - 1][i] -
            star[x1 - 1][y2][i];
      ans += (cnt * light);
    }
    cout << ans << endl;
  }
  return 0;
}
