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.
import java.io.*;

public class starsky {
	static class star{
		int x, y, s;
		public star(int x, int y, int s) {
			this.x = x;
			this.y = y;
			this.s = s;
		}
	}
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] line = br.readLine().split(" ");
		int n = Integer.parseInt(line[0]);
		int q = Integer.parseInt(line[1]);
		int c = Integer.parseInt(line[2]);
		int[][] stars = new int[100][100];
		star[] sl = new star[n];
		int[][][] psums = new int[c+1][101][101];
		for (int i = 0; i < n; i++) {
			line = br.readLine().split(" ");
			stars[Integer.parseInt(line[0])-1][Integer.parseInt(line[1])-1] += Integer.parseInt(line[2]);
			sl[i] = new star(Integer.parseInt(line[0])-1, Integer.parseInt(line[1])-1, Integer.parseInt(line[2]));
		}
		for (int i = 0; i <= c; i++) {
			int sum = 0;
			for (int k = 1; k <= 100; k++) {
				sum += stars[0][k-1];
				psums[i][1][k] = sum;
			}
			for (int j = 2; j <= 100; j++) {
				sum = 0;
				for (int k = 1; k <= 100; k++) {
					sum += stars[j-1][k-1];
					psums[i][j][k] = sum + psums[i][j-1][k];
				}
			}
			for (int j = 0; j < n; j++) {
				int temp = sl[j].s;
				sl[j].s = (sl[j].s+1) % (c+1);
				stars[sl[j].x][sl[j].y] += (sl[j].s - temp);
			}
		}
		for (int i = 0; i < q; i++) {
			line = br.readLine().split(" ");
			int t = Integer.parseInt(line[0]) % (c+1);
			int x1 = Integer.parseInt(line[1])-1;
			int x2 = Integer.parseInt(line[3]);
			int y1 = Integer.parseInt(line[2])-1;
			int y2 = Integer.parseInt(line[4]);
			System.out.println(psums[t][x2][y2] - psums[t][x2][y1] - psums[t][x1][y2] + psums[t][x1][y1]);
		}
	}
}
