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;
template <typename T>
void printvec(vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++) {
    cout << vec[i] << " ";
  }
  cout << "\n";
}
void updateBIT(int BIT[][102 + 1], int x, int y, int val) {
  int i, j;
  for (i = x + 1; i <= 102; i += (i & (-1 * i))) {
    for (j = y + 1; j <= 102; j += (j & (-1 * j))) BIT[i][j] += val;
  }
  return;
}
int getSum(int BIT[][102 + 1], int x, int y) {
  int sum = 0;
  int i, j;
  for (i = x + 1; i > 0; i -= (i & (-1 * i))) {
    for (j = y + 1; j > 0; j -= (j & (-1 * j))) {
      sum += BIT[i][j];
    }
  }
  return sum;
}
int main() {
  int mat[11][102][102], BIT[11][102 + 1][102 + 1];
  int n, q, c;
  cin >> n >> q >> c;
  int i, j, k;
  for (i = 0; i < 11; i++) {
    for (j = 0; j < 102; j++) {
      for (k = 0; k < 102; k++) {
        mat[i][j][k] = 0;
      }
    }
  }
  for (i = 0; i < 11; i++) {
    for (j = 0; j < 102 + 1; j++) {
      for (k = 0; k < 102 + 1; k++) {
        BIT[i][j][k] = 0;
      }
    }
  }
  for (i = 0; i < n; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    mat[0][x][y] += z;
    updateBIT(BIT[0], x, y, z);
    int pre = z;
    for (j = 1; j < 11; j++) {
      pre++;
      if (pre > c) pre = 0;
      mat[j][x][y] += (pre);
      updateBIT(BIT[j], x, y, pre);
    }
  }
  for (i = 0; i < 102 + 1; i++) {
    for (j = 0; j < 102 + 1; j++) {
    }
  }
  for (int iii = 0; iii < q; iii++) {
    int ti, x1, y1, x2, y2;
    cin >> ti >> x1 >> y1 >> x2 >> y2;
    int ans = 0;
    ti = (ti % (c + 1));
    if (ti <= 10) {
      ans = getSum(BIT[ti], x2, y2) - getSum(BIT[ti], x2, y1 - 1) -
            getSum(BIT[ti], x1 - 1, y2) + getSum(BIT[ti], x1 - 1, y1 - 1);
    } else {
      ans = getSum(BIT[10], x2, y2) - getSum(BIT[10], x2, y1 - 1) -
            getSum(BIT[10], x1 - 1, y2) + getSum(BIT[10], x1 - 1, y1 - 1);
    }
    cout << ans << "\n";
  }
  return (0);
}
