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 long long INF = 9e18;
const int N = 1e2 + 5;
struct BIT {
  int a[N][N];
  int n;
  void init(int _n) {
    n = _n;
    memset(a, 0, sizeof a);
  }
  void add(int x, int y, int val) {
    for (int i = x; i <= n; i += i & -i) {
      for (int j = y; j <= n; j += j & -j) a[i][j] += val;
    }
  }
  int sum(int x, int y) {
    int ret = 0;
    for (int i = x; i > 0; i -= i & -i) {
      for (int j = y; j > 0; j -= j & -j) ret += a[i][j];
    }
    return ret;
  }
  int sum(int x1, int y1, int x2, int y2) {
    return sum(x2, y2) - sum(x1 - 1, y2) - sum(x2, y1 - 1) +
           sum(x1 - 1, y1 - 1);
  }
} bit[11];
int main() {
  int n, q, c;
  scanf("%d%d%d", &n, &q, &c);
  for (int i = 0; i <= c; i++) bit[i].init(100);
  for (int i = 0; i < n; i++) {
    int x, y, s;
    scanf("%d%d%d", &x, &y, &s);
    bit[s].add(x, y, 1);
  }
  while (q--) {
    int t, x1, y1, x2, y2;
    scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2);
    int ans = 0;
    for (int i = 0; i <= c; i++) {
      ans += (i + t) % (c + 1) * bit[i].sum(x1, y1, x2, y2);
    }
    printf("%d\n", ans);
  }
  return 0;
}
