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 dp[20][110][110], g[110][110];
struct bkn {
  int x, y, s;
} a[100010];
int main() {
  int n, t, x1, x2, y1, y2, q, c, i, j, k;
  cin >> n >> q >> c;
  for (i = 1; i <= n; i++) {
    cin >> a[i].x >> a[i].y >> a[i].s;
    a[i].s--;
  }
  for (i = 0; i <= c; i++) {
    memset(g, 0, sizeof(g));
    for (j = 1; j <= n; j++) {
      a[j].s++;
      if (a[j].s > c) a[j].s = 0;
      g[a[j].x][a[j].y] += a[j].s;
    }
    for (j = 1; j <= 100; j++)
      for (k = 1; k <= 100; k++) {
        dp[i][j][k] =
            dp[i][j - 1][k] + dp[i][j][k - 1] - dp[i][j - 1][k - 1] + g[j][k];
      }
  }
  while (q--) {
    cin >> t >> x1 >> y1 >> x2 >> y2;
    t %= (c + 1);
    cout << dp[t][x2][y2] - dp[t][x1 - 1][y2] - dp[t][x2][y1 - 1] +
                dp[t][x1 - 1][y1 - 1]
         << endl;
  }
}
