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>
#pragma GCC optimize("O3")
using namespace std;
const long long mod = (1e+9) + 7;
const long long sz = 2e5 + 9;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long a, long long b, long long m = mod) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    b = b >> 1;
    a = (a * a) % m;
  }
  return res;
}
long long dp[11][101][101];
void solve() {
  long long n, q, c;
  cin >> n >> q >> c;
  long long x, y, si;
  for (long long i = 0; i < n; i++) cin >> x >> y >> si, dp[si][x][y]++;
  for (long long p = 0; p <= c; p++) {
    for (long long i = 1; i <= 100; i++)
      for (long long j = 1; j <= 100; j++)
        dp[p][i][j] += dp[p][i - 1][j] + dp[p][i][j - 1] - dp[p][i - 1][j - 1];
  }
  long long t, x1, y1, x2, y2;
  while (q--) {
    cin >> t >> x1 >> y1 >> x2 >> y2;
    long long ans = 0;
    for (long long p = 0; p <= c; p++) {
      long long brightness = (p + t) % (c + 1);
      long long stars = dp[p][x2][y2] - dp[p][x1 - 1][y2] - dp[p][x2][y1 - 1] +
                        dp[p][x1 - 1][y1 - 1];
      ans += stars * brightness;
    }
    cout << ans << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t beg = clock();
  solve();
  clock_t end = clock();
  fprintf(stderr, "%.3f sec, Copyright %c 2019 Skyscraper. \n",
          double(end - beg) / CLOCKS_PER_SEC, 184);
  return 0;
}
