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;
int x[100005], y[100005];
map<pair<int, int>, int> mapi, val, mapii;
vector<vector<int> > vec(100005);
long long sum[105][105], counti[105][105][20][2];
int main() {
  int n, q, c, i, j, k, s, x1;
  cin >> n >> q >> c;
  k = 1;
  for (i = 0; i < 101; i++) {
    for (j = 0; j < 101; j++) {
      mapii[make_pair(i, j)] = k;
      k++;
    }
  }
  for (i = 0; i < n; i++) {
    scanf("%d%d%d", &x[i], &y[i], &s);
    mapi[make_pair(x[i], y[i])] = 1;
    val[make_pair(x[i], y[i])] += s;
    vec[mapii[make_pair(x[i], y[i])]].push_back(s);
  }
  for (k = 0; k < 11; k++) {
    for (i = 1; i < 101; i++) {
      for (j = 1; j < 101; j++) {
        sum[i][j] = sum[i][j - 1] + val[make_pair(i, j)];
        counti[i][j][k][1] = counti[i][j - 1][k][1];
        counti[i][j][k][0] = counti[i][j - 1][k][0];
        if (mapi[make_pair(i, j)] == 1) {
          for (x1 = 0; x1 < vec[mapii[make_pair(i, j)]].size(); x1++) {
            if (vec[mapii[make_pair(i, j)]][x1] + k <= c) {
              counti[i][j][k][0]++;
            } else {
              counti[i][j][k][1]++;
            }
          }
        }
      }
    }
  }
  for (j = 0; j < q; j++) {
    long long sumi = 0, val0 = 0, val1 = 0;
    int x1, y1, x2, y2, t;
    scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2);
    t %= (c + 1);
    for (i = x1; i < x2 + 1; i++) {
      val0 += counti[i][y2][t][0] - counti[i][y1 - 1][t][0];
      sumi += sum[i][y2] - sum[i][y1 - 1];
      val1 += counti[i][y2][t][1] - counti[i][y1 - 1][t][1];
    }
    val0 *= t;
    val1 *= (t - c - 1);
    sumi += val0;
    sumi += val1;
    cout << sumi << endl;
  }
}
