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 int read() {
  int X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return w ? -X : X;
}
int n, q, c;
int dp[(100 + 10)][(100 + 10)][20];
class star {
 public:
  int x, y, s;
  void init(int i) {
    x = read(), y = read(), s = read();
    dp[x][y][s]++;
  }
} a[(100000 + 100)];
int calc(int s, int t, int n) { return (s + t) % (c + 1) * n; }
void first() {
  for (int i = 1; i < (100 + 10); i++) {
    for (int j = 1; j < (100 + 10); j++) {
      for (int k = 0; k <= c; k++) {
        dp[i][j][k] += dp[i - 1][j][k] + dp[i][j - 1][k] - dp[i - 1][j - 1][k];
      }
    }
  }
}
long long findAns(int x1, int y1, int x2, int y2, int t) {
  long long ans = 0;
  for (int i = 0; i <= c; i++) {
    int num = dp[x2][y2][i] - dp[x2][y1 - 1][i] - dp[x1 - 1][y2][i] +
              dp[x1 - 1][y1 - 1][i];
    ans += calc(i, t, num);
  }
  return ans;
}
int main() {
  n = read(), q = read(), c = read();
  for (int i = 1; i <= n; i++) a[i].init(i);
  first();
  for (int i = 1; i <= q; i++) {
    int t = read(), x1 = read(), y1 = read(), x2 = read(), y2 = read();
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y2, y2);
    cout << findAns(x1, y1, x2, y2, t) << endl;
  }
  return 0;
}
