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 = 105, M = 15;
int n, q, c, x, y, t, ans, xx, yy;
int add[N][N][M], sum[N][N][M];
int main() {
  scanf("%d%d%d", &n, &q, &c);
  for (register int i = 1; i <= n; ++i)
    scanf("%d%d%d", &x, &y, &t), add[x][y][t]++;
  for (register int k = 0; k <= c; ++k) {
    for (register int i = 1; i <= 100; ++i)
      for (register int j = 1; j <= 100; ++j)
        sum[i][j][k] = sum[i][j - 1][k] + sum[i - 1][j][k] -
                       sum[i - 1][j - 1][k] + add[i][j][k];
  }
  while (q--) {
    ans = 0;
    scanf("%d%d%d%d%d", &t, &x, &y, &xx, &yy);
    for (register int i = 0; i <= c; ++i) {
      int k = (t + i) % (c + 1);
      ans += k * (sum[xx][yy][i] - sum[xx][y - 1][i] - sum[x - 1][yy][i] +
                  sum[x - 1][y - 1][i]);
    }
    printf("%d\n", ans);
  }
  return 0;
}
