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 maxn = 1e5 + 3;
int n, q, c;
long long a[13][103][103], sum[13][103][103];
int get(int x, int r1, int c1, int r2, int c2) {
  return sum[x][r2][c2] - sum[x][r1 - 1][c2] - sum[x][r2][c1 - 1] +
         sum[x][r1 - 1][c1 - 1];
}
int main() {
  memset(a, 0, sizeof(a));
  memset(sum, 0, sizeof(sum));
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> q >> c;
  for (int i = 0; i < n; ++i) {
    int x, y, s;
    cin >> x >> y >> s;
    a[s][x][y]++;
  }
  for (int x = 0; x <= 10; x++) {
    for (int i = 1; i <= 102; ++i)
      for (int j = 1; j <= 102; ++j) {
        sum[x][i][j] = a[x][i][j] + sum[x][i - 1][j] + sum[x][i][j - 1] -
                       sum[x][i - 1][j - 1];
      }
  }
  while (q--) {
    int t, x1, x2, y1, y2;
    cin >> t >> x1 >> y1 >> x2 >> y2;
    long long ans = 0;
    for (int i = 0; i <= c; i++) {
      ans += 1LL * get(i, x1, y1, x2, y2) * ((i + t) % (c + 1));
    }
    cout << ans << '\n';
  }
  return 0;
}
