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 bool eq(double a, double b) { return fabs(a - b) < 1e-9; }
const int inf = 1 << 29;
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
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); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
int dp[104][104][12];
long long int t;
int x1, y111, x2, y2;
void solve() {
  for (int i = (1); i < (104); i++) {
    for (int j = (1); j < (104); j++) {
      for (int k = (0); k < (12); k++) {
        dp[i][j][k] += dp[i][j - 1][k];
      }
    }
  }
  for (int j = (1); j < (104); j++) {
    for (int i = (1); i < (104); i++) {
      for (int k = (0); k < (12); k++) {
        dp[i][j][k] += dp[i - 1][j][k];
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, q, c;
  cin >> n >> q >> c;
  for (int i = 0; i < (n); i++) {
    int a, b, c;
    cin >> a >> b >> c;
    dp[a][b][c] += 1;
  }
  solve();
  for (int i = 0; i < (q); i++) {
    cin >> t >> x1 >> y111 >> x2 >> y2;
    int second[12] = {0};
    long long int ans = 0;
    for (int j = (0); j < (12); j++) {
      second[j] = dp[x2][y2][j] - dp[x1 - 1][y2][j] - dp[x2][y111 - 1][j] +
                  dp[x1 - 1][y111 - 1][j];
      ans += second[j] * ((j + t) % (c + 1));
    }
    cout << ans << endl;
  }
  return 0;
}
