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 MOD = 1e9 + 7;
const long double PI = 4 * atan((long double)1);
long long power(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res *= a;
    a *= a;
    n /= 2;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n, q, c;
  cin >> n >> q >> c;
  vector<pair<long long, long long> > vx[101];
  vector<long long> vx1[101];
  for (int i = (0); i < (n); ++i) {
    long long int x, y, c1;
    cin >> x >> y >> c1;
    vx[x].push_back({y, c1});
  }
  for (int i = (0); i < (101); ++i) {
    sort(vx[i].begin(), vx[i].end());
    if (vx[i].size() != 0)
      for (int j = (0); j < (vx[i].size()); ++j)
        vx1[i].push_back(vx[i][j].first);
  }
  vector<long long> dp[c + 1][101];
  for (int k = (0); k < (c + 1); ++k) {
    for (int i = (0); i < (101); ++i) {
      dp[k][i].push_back(0);
      if (vx[i].size() != 0)
        for (int j = (0); j < (vx[i].size()); ++j)
          dp[k][i].push_back(dp[k][i][j] + (vx[i][j].second + k) % (c + 1));
    }
  }
  for (int i = (0); i < (q); ++i) {
    long long int c2, x1, y1, x2, y2, ans = 0;
    cin >> c2 >> x1 >> y1 >> x2 >> y2;
    c2 = c2 % (c + 1);
    for (int i = (x1); i < (x2 + 1); ++i) {
      long long int ind1 =
          lower_bound(vx1[i].begin(), vx1[i].end(), y1) - vx1[i].begin();
      long long int ind2 =
          upper_bound(vx1[i].begin(), vx1[i].end(), y2) - vx1[i].begin();
      ans += (dp[c2][i][ind2] - dp[c2][i][ind1]);
    }
    cout << ans;
    cout << "\n";
  }
  return 0;
}
