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.util.*;
import java.io.*;
public class star {
	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		// BufferedReader in = new BufferedReader(new FileReader("test.in"));
		PrintStream out = System.out;
		String line = in.readLine().trim(), lines[];
		lines = line.split(" +");
		int n = Integer.parseInt(lines[0]), q = Integer.parseInt(lines[1]), c = Integer.parseInt(lines[2]);

		int[][][] mat = new int[101][101][c + 1];
		for (int i = 0; i < n; i++) {
			lines = in.readLine().split(" +");
			mat[Integer.parseInt(lines[0])][Integer.parseInt(lines[1])][Integer.parseInt(lines[2])]++;
		}

		int[][][] pre = new int[101][101][c + 1];
		for (int k = 0; k < c + 1; k++) pre[0][0][k] = mat[0][0][k];
		for (int i = 1; i <= 100; i++) {
			for (int j = 1; j <= 100; j++) {
				for (int k = 0; k < c + 1; k++) {
					pre[i][j][k] = pre[i - 1][j][k] + pre[i][j - 1][k] - pre[i - 1][j - 1][k] + mat[i][j][k];
				}
			}
		}

		int[][] views = new int[q][5];
		for (int i = 0; i < q; i++) {
			lines = in.readLine().split(" +");
			for (int j = 0; j < 5; j++) views[i][j] = Integer.parseInt(lines[j]);
		}
		for (int qq = 0; qq < q; qq++) {
			int sum = 0;
			for (int k = 0; k <= c; k++) {
				sum += (k + views[qq][0]) % (c + 1) * pre[views[qq][3]][views[qq][4]][k];
			}
			for (int k = 0; k <= c; k++) {
				sum -= (k + views[qq][0]) % (c + 1) * pre[views[qq][1] - 1][views[qq][4]][k];
				sum -= (k + views[qq][0]) % (c + 1) * pre[views[qq][3]][views[qq][2] - 1][k];
			}
			for (int k = 0; k <= c; k++) {
				sum += (k + views[qq][0]) % (c + 1) * pre[views[qq][1] - 1][views[qq][2] - 1][k];
			}
			out.println(sum);
		}
	}
}