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;
class CStarSky {
  int N, Q, C;
  vector<vector<vector<int>>> S;

 public:
  void input(std::istream& in, std::ostream& out) {
    ios_base::sync_with_stdio(false);
    in >> N >> Q >> C;
    S = vector<vector<vector<int>>>(
        C + 1, vector<vector<int>>(110, vector<int>(110, 0)));
    for (int(i) = (0); (i) < (N); ++(i)) {
      int x, y, b;
      in >> x >> y >> b;
      for (int(j) = (0); (j) < (C + 1); ++(j))
        S[j][x + 1][y + 1] += (b + j) % (C + 1);
    }
  }
  void query(std::istream& in, std::ostream& out) {
    int t, x1, y1, x2, y2;
    in >> t >> x1 >> y1 >> x2 >> y2;
    int res = S[t % (C + 1)][x2 + 1][y2 + 1] - S[t % (C + 1)][x2 + 1][y1] -
              S[t % (C + 1)][x1][y2 + 1] + S[t % (C + 1)][x1][y1];
    out << res << '\n';
  }
  void solve(std::istream& in, std::ostream& out) {
    input(in, out);
    for (int(i) = (0); (i) < (C + 1); ++(i)) {
      for (int(j) = (1); (j) < (101 + 1); ++(j)) {
        for (int(k) = (1); (k) < (100 + 1); ++(k)) S[i][j][k + 1] += S[i][j][k];
        for (int(k) = (1); (k) < (101 + 1); ++(k)) S[i][j][k] += S[i][j - 1][k];
      }
    }
    for (int(i) = (0); (i) < (Q); ++(i)) {
      query(in, out);
    }
  }
};
int main() {
  CStarSky solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
