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 Main {
	
	private static int[][][] cnt = new int[200][200][30];
	
	private static void build() {
		
		int i , j , k;
		for (k = 0;k <= 10;k ++) {
			for (i = 1;i <= 100;i ++) {
				if (i > 1) {
					cnt[i][1][k] += cnt[i - 1][1][k];
				}
			}
			for (j = 1;j <= 100;j ++) {
				if (j > 1) {
					cnt[1][j][k] += cnt[1][j - 1][k];
				}
			}
			for (i = 2;i <= 100;i ++) {
				for (j = 2;j <= 100;j ++) {
					int temp = cnt[i - 1][j][k] + cnt[i][j - 1][k] - cnt[i - 1][j - 1][k] + cnt[i][j][k];
					cnt[i][j][k] = temp;
				}
			}
		}
		
	}

	public static void main(String[] args) {
		
		Scanner scan = new Scanner(System.in);
		int i , j , n , q , c;
		n = scan.nextInt();
		q = scan.nextInt();
		c = scan.nextInt();
		for (i = 0;i < n;i ++) {
			int x , y , s;
			x = scan.nextInt();
			y = scan.nextInt();
			s = scan.nextInt();
			cnt[x][y][s] ++;
		}
		build();
		for (i = 0;i < q;i ++) {
			int t , x1 , y1 , x2 , y2;
			t = scan.nextInt();
			x1 = scan.nextInt();
			y1 = scan.nextInt();
			x2 = scan.nextInt();
			y2 = scan.nextInt();
			int ans = 0;
			for (j = 0;j <= c;j ++) {
				int bright = (j + t) % (c + 1);
				int temp = cnt[x2][y2][j] + cnt[x1 - 1][y1 - 1][j] - cnt[x1 - 1][y2][j] - cnt[x2][y1 - 1][j];
				ans += temp * bright;
			}
			System.out.println(ans);
		}
		
    }
    
}

















