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.ArrayList;
import java.util.Scanner;

public class CStarSky {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int q = in.nextInt();
		int c = in.nextInt();
		
		int[][][] brightness = new int[101][101][c+1];
		for (int i = 0; i < n; i++) {
			int x = in.nextInt();
			int y = in.nextInt();
			int bright = in.nextInt();
			
			for( int t = 0; t <= c; t++ ) {
				brightness[x][y][t] += (bright + t) % (c+1);
			}
		}
		
		int[][][] prefixBrightness = new int[101][101][c+1];
		for( int x = 0; x <= 100; x++ ) {
			for( int y = 0; y <= 100; y++ ) {
				for( int t = 0; t <= c; t++ ) {
					if( y == 0 )
						prefixBrightness[x][y][t] = brightness[x][y][t];
					else
						prefixBrightness[x][y][t] = brightness[x][y][t] + prefixBrightness[x][y-1][t];
				}
			}
		}

		for (int i = 0; i < q; i++) {
			int time = in.nextInt() % (c+1);
			int x1 = in.nextInt();
			int y1 = in.nextInt();
			int x2 = in.nextInt();
			int y2 = in.nextInt();
		
			long totalBrightness = 0;

			for (int x = x1; x <= x2; x++) {
				if( y1 == 0 )
					totalBrightness += prefixBrightness[x][y2][time];
				else
					totalBrightness += prefixBrightness[x][y2][time] - prefixBrightness[x][y1-1][time];
			}

			System.out.println(totalBrightness);
		}
		
		in.close();
	}
}