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 int MOD = 1e9 + 7;
const int N = 1e6 + 5;
long long int POWER[65];
long long int aonb(long long int a, long long int b) {
  long long int ret = 1;
  while (b) {
    if (b & 1) ret *= a;
    a *= a;
    if (ret >= MOD) ret %= MOD;
    if (a >= MOD) a %= MOD;
    b >>= 1;
  }
  return ret;
}
void precompute() {
  POWER[0] = 1;
  for (int i = 1; i < 63; i++) POWER[i] = POWER[i - 1] << 1LL;
}
long long int dp[11][101][101];
int main() {
  long long int A[101][101];
  int count[101][101];
  for (int i = 0; i < 101; i++) {
    for (int j = 0; j < 101; j++) {
      A[i][j] = 0;
      count[i][j] = 0;
    }
  }
  int n, que, c;
  int x, y, b;
  scanf("%d %d %d", &n, &que, &c);
  for (int i = 0; i < n; i++) {
    scanf("%d %d %d", &x, &y, &b);
    A[x][y] += b;
    count[x][y]++;
    dp[0][x][y] += b;
    for (int ch = 1; ch <= c; ch++) {
      b = (b + 1) % (c + 1);
      dp[ch][x][y] += b;
    }
  }
  for (int ch = 0; ch <= c; ch++) {
    for (int i = 1; i < 101; i++) {
      for (int j = 1; j < 101; j++) {
        dp[ch][i][j] +=
            dp[ch][i - 1][j] + dp[ch][i][j - 1] - dp[ch][i - 1][j - 1];
      }
    }
  }
  long long int t;
  int p, q;
  long long int ans;
  while (que--) {
    scanf("%lld %d %d %d %d", &t, &x, &y, &p, &q);
    t = t % (c + 1);
    ans = dp[t][p][q];
    if (y > 1) ans -= dp[t][p][y - 1];
    if (x > 1) ans -= dp[t][x - 1][q];
    if (x > 1 && y > 1) ans += dp[t][x - 1][y - 1];
    printf("%lld\n", ans);
  }
  return 0;
}
