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 = 107;
vector<int> g[N][N];
int par[12][N][N];
int n, q, c;
int que(int num, int x1, int y1, int x2, int y2) {
  x1--;
  y1--;
  return par[num][x2][y2] - par[num][x1][y2] - par[num][x2][y1] +
         par[num][x1][y1];
}
int sum(int x, int y) {
  int s = 0;
  for (int k = 0; k < g[x][y].size(); k++) s += g[x][y][k];
  return s;
}
void prt() {
  int k;
  for (int i = 1; i <= 5; i++, printf("\n")) {
    for (int j = 1; j <= 5; j++) {
      for (k = 0; k < g[i][j].size(); k++) printf("%3d", g[i][j][k]);
      if (!k) printf("%3d", -1);
    }
  }
  printf("\n");
}
int main() {
  int vis = 1;
  scanf("%d%d%d", &n, &q, &c);
  int x, y, s, x1, y1, x2, y2, t;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d", &x, &y, &s);
    g[x][y].push_back(s);
  }
  for (int i = 1; i <= 100; i++)
    for (int j = 1; j <= 100; j++)
      par[0][i][j] = par[0][i - 1][j] + par[0][i][j - 1] -
                     par[0][i - 1][j - 1] + sum(i, j);
  for (int l = 1; l <= 10; l++) {
    for (int i = 1; i <= 100; i++) {
      for (int j = 1; j <= 100; j++) {
        for (int k = 0; k < g[i][j].size(); k++) {
          if (g[i][j][k] < c)
            g[i][j][k]++;
          else
            g[i][j][k] = 0;
        }
      }
    }
    for (int i = 1; i <= 100; i++)
      for (int j = 1; j <= 100; j++)
        par[l][i][j] = par[l][i - 1][j] + par[l][i][j - 1] -
                       par[l][i - 1][j - 1] + sum(i, j);
  }
  bool ok = false;
  for (int i = 0; i < q; i++) {
    scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2);
    printf("%d\n", que(t % (c + 1), x1, y1, x2, y2));
  }
  return 0;
}
