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>
std::vector<std::vector<std::vector<int>>> space;
int c;
int findBrInDot(int x, int y, int curTime) {
  int sum = 0;
  for (auto starBr : space[x][y]) {
    sum += (starBr + curTime) % (c + 1);
  }
  return sum;
}
int main() {
  int n, q;
  scanf("%d %d %d", &n, &q, &c);
  space.resize(100);
  for (int i = 0; i < 100; ++i) {
    space[i].resize(100);
  }
  for (int i = 0; i < n; ++i) {
    int x, y, s;
    scanf("%d %d %d", &x, &y, &s);
    space[x - 1][y - 1].push_back(s);
  }
  std::vector<std::vector<std::vector<int>>> lines(
      c + 1, std::vector<std::vector<int>>(100, std::vector<int>(100)));
  for (int curT = 0; curT <= c; ++curT) {
    for (int i = 0; i < 100; ++i) {
      lines[curT][i][0] = findBrInDot(i, 0, curT);
    }
    for (int i = 0; i < 100; ++i) {
      for (int j = 1; j < 100; ++j) {
        lines[curT][i][j] = lines[curT][i][j - 1] + findBrInDot(i, j, curT);
      }
    }
  }
  for (int curT = 0; curT <= c; ++curT) {
    for (int i = 1; i < 100; ++i) {
      for (int j = 0; j < 100; ++j) {
        lines[curT][i][j] += lines[curT][i - 1][j];
      }
    }
  }
  int x1, x2, y1, y2, time;
  for (int i = 0; i < q; ++i) {
    scanf("%d %d %d %d %d", &time, &x1, &y1, &x2, &y2);
    --x1;
    --y1;
    --x2;
    --y2;
    time %= (c + 1);
    long long ans = 0;
    ans += (long long)lines[time][x2][y2];
    if (x1 > 0) ans -= (long long)lines[time][x1 - 1][y2];
    if (y1 > 0) ans -= (long long)lines[time][x2][y1 - 1];
    if (x1 > 0 && y1 > 0) ans += (long long)lines[time][x1 - 1][y1 - 1];
    std::cout << ans << "\n";
  }
}
