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;
inline int two(int n) { return 1 << n; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
const int p = 11;
const int c = 110;
int arr[p][c][c];
int n, mod, q;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q >> mod;
  for (int i = 0; i < n; ++i) {
    int a, b, s;
    cin >> a >> b >> s;
    arr[s][a][b]++;
  }
  for (int i = 0; i <= mod; ++i) {
    for (int j = 1; j <= c; ++j) {
      for (int k = 1; k <= c; ++k) {
        arr[i][j][k] +=
            arr[i][j - 1][k] + arr[i][j][k - 1] - arr[i][j - 1][k - 1];
      }
    }
  }
  for (int i = 0; i < q; ++i) {
    int t, x1, y1, x2, y2;
    cin >> t >> x1 >> y1 >> x2 >> y2;
    int ans = 0;
    for (int j = 0; j <= mod; ++j) {
      int bright = (j + t) % (mod + 1);
      int amount = arr[j][x2][y2] - arr[j][x1 - 1][y2] - arr[j][x2][y1 - 1] +
                   arr[j][x1 - 1][y1 - 1];
      ans += (amount * bright);
    }
    cout << ans << '\n';
  }
}
