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;
long long int fact(long long int n) {
  if (n == 1 || n == 0) return 1;
  return n * fact(n - 1);
}
int isPrime(long long int n) {
  long long int i, j;
  for (i = 2; i < sqrt(n) + 1; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long int modexpo(long long int x, long long int p) {
  long long int res = 1;
  x = x % 1000000007;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % 1000000007;
    res %= 1000000007;
  }
  return res;
}
long long int binSearch(long long int a[], long long int l, long long int h,
                        long long int x) {
  long long int c = -1;
  while (l <= h) {
    long long int m = (l + h) / 2;
    if (a[m] >= x) {
      c = m;
      h = m - 1;
    } else
      l = m + 1;
  }
  return c;
}
void pairsort(long long int a[], long long int b[], long long int n) {
  pair<long long int, long long int> pairt[n];
  for (long long int i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long int i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long int nod(long long int n) {
  long long int c = 0;
  while (n > 0) {
    c++;
    n /= 10;
  }
  return c;
}
long long int sod(long long int n) {
  long long int s = 0;
  while (n > 0) {
    s += n % 10;
    n /= 10;
  }
  return s;
}
long long int a[101][101][20];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int q, c, n, m, k = 1, i, j, f = 0, g = 0, h = 1, sum = 0, ans = 0,
                            x, y, l;
  cin >> n >> q >> c;
  for (i = 0; i < n; i++) {
    cin >> x >> y >> f;
    a[x][y][f]++;
  }
  for (i = 1; i < 101; i++) {
    for (j = 1; j < 101; j++) {
      for (k = 0; k < c + 1; k++) {
        a[i][j][k] += a[i][j - 1][k] + a[i - 1][j][k] - a[i - 1][j - 1][k];
      }
    }
  }
  long long int x1, x2, y1, y2, t;
  while (q--) {
    cin >> t >> x1 >> y1 >> x2 >> y2;
    sum = 0;
    for (i = 0; i < c + 1; i++) {
      f = (t + i) % (c + 1);
      long long int cnt = a[x2][y2][i] - (a[x1 - 1][y2][i] + a[x2][y1 - 1][i] -
                                          a[x1 - 1][y1 - 1][i]);
      sum += f * cnt;
    }
    cout << sum << endl;
  }
  return 0;
}
