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.Scanner;

public class Codeforces835C {
	
	static int n ,c , q;
	static int precomp[][][];
	
	public static void main(String[] args) {
		solve();
	}
	static void solve() {
		Scanner sc = new Scanner(System.in);
		
		n = sc.nextInt();
		q = sc.nextInt();
		c = sc.nextInt();
		
		precomp = new int[c + 1][101][101];
		for(int i = 0 ; i < n ;i++) {
			int x = sc.nextInt(), y= sc.nextInt(),s = sc.nextInt();
			precomp[s][x][y]++;
		}
		// computing 3d prefix sum for easy calculation
		for(int i = 0 ; i <= c  ; i++) {
			for(int x = 1 ; x <= 100 ; x++) {
				for(int y = 1 ; y <= 100 ; y++ ) {
					precomp[i][x][y] += precomp[i][x-1][y] + precomp[i][x][y-1] - precomp[i][x-1][y-1];
				}
			}
		}
		StringBuilder sb = new StringBuilder();
		for(int i = 0 ; i < q ; i++) {
			int t = sc.nextInt();
			int x1 = sc.nextInt();
            int y1 = sc.nextInt();
            int x2 = sc.nextInt();
            int y2 = sc.nextInt();
            sb.append(get(t,x1,y1,x2,y2) + "\n");
		}
		System.out.println(sb);
		sc.close();
	}
	static int get(int t, int x1, int y1, int x2, int y2) {
		int ans = 0;
        for (int i = 0; i <= c; i++) {
            int brightness = (i + t) % (c + 1);
            int amount = precomp[i][x2][y2] - precomp[i][x1 - 1][y2] - precomp[i][x2][y1 - 1] + precomp[i][x1 - 1][y1 - 1];
            ans += amount * brightness;
        }
        return ans;
	}

}