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;
vector<long long> mat[101][101];
bool star[101][101];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, q, c;
  cin >> n >> q >> c;
  memset(star, 0, sizeof(star));
  memset(mat, 0, sizeof(mat));
  for (int i = 0; i < (int)(n); i++) {
    int x, y, second;
    cin >> x >> y >> second;
    star[x][y] = true;
    mat[x][y].push_back(second);
  }
  vector<vector<tuple<int, int, int, int, int>>> queries(11);
  vector<pair<int, long long>> res;
  for (int i = 0; i < (int)(q); i++) {
    int t, x0, y0, x1, y1;
    cin >> t >> x0 >> y0 >> x1 >> y1;
    t = t % (c + 1);
    queries[t].push_back(make_tuple(x0, y0, x1, y1, i));
  }
  for (int b = 0; b <= c; b++) {
    vector<vector<long long>> acum(101, vector<long long>(101, 0));
    for (int x = 1; x <= (int)(100); x++) {
      acum[x][0] = 0;
      acum[x][1] = 0;
      for (auto val : mat[x][1]) {
        acum[x][1] += val;
      }
      for (int y = 2; y <= (int)(100); y++) {
        acum[x][y] = acum[x][y - 1];
        for (auto val : mat[x][y]) {
          acum[x][y] += val;
        }
      }
    }
    for (auto q : queries[b]) {
      long long soma = 0;
      int x0, y0, x1, y1, idx;
      tie(x0, y0, x1, y1, idx) = q;
      for (int x = x0; x <= x1; x++) {
        soma += acum[x][y1];
        soma -= acum[x][y0 - 1];
      }
      res.push_back(make_pair(idx, soma));
    }
    for (int x = 1; x <= (int)(100); x++) {
      for (int y = 1; y <= (int)(100); y++) {
        if (mat[x][y].size()) {
          for (int i = 0; i < (int)mat[x][y].size(); i++) {
            mat[x][y][i] = (mat[x][y][i] + 1) % (c + 1);
          }
        }
      }
    }
  }
  sort(res.begin(), res.end());
  for (auto r : res) {
    cout << r.second << "\n";
  }
  return 0;
}
