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>
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;
}
long long int gcd(long long int a, long long int b) {
  return (a ? gcd(b % a, a) : b);
}
long long int modPow(long long int a, long long int b, long long int MOD) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % MOD;
    }
    b /= 2;
    y = (y * y) % MOD;
  }
  return x;
}
long long int modInverse(long long int a, long long int p) {
  return modPow(a, p - 2, p);
}
using namespace std;
const int N = 103;
long long int BIT[N][N][N];
void updateBIT(long long int x, long long int y, long long int b,
               long long int val) {
  for (; x <= N; x += x & -x) {
    for (long long int y1 = y; y1 <= N; y1 += y1 & -y1) BIT[x][y1][b] += val;
  }
}
long long int queryBIT(long long int x, long long int y, long long int b) {
  long long int sum = 0;
  for (; x > 0; x -= x & -x) {
    for (long long int y1 = y; y1 > 0; y1 -= y1 & -y1) sum += BIT[x][y1][b];
  }
  return sum;
}
int main() {
  long long int n, q, c, i, j, k;
  cin >> n >> q >> c;
  for (int i = 0; i < (n); i++) {
    long long int x, y, b;
    cin >> x >> y >> b;
    updateBIT(x, y, b, 1);
  }
  while (q--) {
    long long int t1, x1, y1, x2, y2;
    cin >> t1 >> x1 >> y1 >> x2 >> y2;
    long long int ans = 0;
    for (int i = (0); i <= (c); i++) {
      long long int temp = (i + t1) % (c + 1);
      long long int cnt = queryBIT(x2, y2, i) - queryBIT(x1 - 1, y2, i) -
                          queryBIT(x2, y1 - 1, i) + queryBIT(x1 - 1, y1 - 1, i);
      ans += cnt * temp;
    }
    cout << ans << endl;
  }
}
