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;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
struct MM {
  long long G[(125)][(125)], S[(125)][(125)];
  void sum(int N, int M) {
    static long long H[(125)][(125)];
    for (int i(0); i < N; i++)
      for (int j(0); j < M; j++) H[i][j] = (i ? H[i - 1][j] : 0) + G[i][j];
    for (int i(0); i < N; i++)
      for (int j(0); j < M; j++) S[i][j] = H[i][j] + (j ? S[i][j - 1] : 0);
  }
  long long sq(int ax, int ay, int bx, int by) {
    long long W(S[bx][by]);
    if (ax) W -= S[ax - 1][by];
    if (ay) W -= S[bx][ay - 1];
    if (ax && ay) W += S[ax - 1][ay - 1];
    return W;
  }
} H[(12)];
int N, Q, C, x, y, X, Y, a;
int main(void) {
  scanf("%d%d%d", &N, &Q, &C);
  for (int i(0); i < N; i++) {
    scanf("%d%d%d", &x, &y, &a);
    for (int i(0); i < (12); i++) H[i].G[x][y] += a, ++a %= C + 1;
  }
  for (int i(0); i < (12); i++) H[i].sum((125) - 2, (125) - 2);
  for (int i(0); i < Q; i++)
    scanf("%d%d%d%d%d", &a, &x, &y, &X, &Y), a %= C + 1,
        printf("%lld\n", H[a].sq(x, y, X, Y));
  return 0;
}
