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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
	static int n,q,c;
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//		BufferedReader br = new BufferedReader(new FileReader("out.txt"));
		StringTokenizer st = new StringTokenizer(br.readLine());
		n = Integer.parseInt(st.nextToken());
		q = Integer.parseInt(st.nextToken());
		c = Integer.parseInt(st.nextToken())+1;
		
		int[][][] count = new int[11][100][100];
		for (int i = 0; i < n; i++) 
		{
			st = new StringTokenizer(br.readLine());
			int x = Integer.parseInt(st.nextToken())-1;
			int y = Integer.parseInt(st.nextToken())-1;
			int c = Integer.parseInt(st.nextToken());
			count[c][y][x] ++;
		}
		
		for (int z = 0; z  < 11; z++) 
			for (int i = 0; i < 100; i++) 
				for (int j = 0; j < 100; j++) 
				{
					if(i>0) count[z][i][j] += count[z][i-1][j];
					if (j > 0) count[z][i][j] += count[z][i][j - 1]; 
					if (i > 0 && j > 0) count[z][i][j] -= count[z][i - 1][j - 1]; 
				}
		
		StringBuilder sb = new StringBuilder();
		for (int z = 0; z < q; z++) 
		{
			st = new StringTokenizer(br.readLine());
			int t = Integer.parseInt(st.nextToken());
			int j = Integer.parseInt(st.nextToken())-1;
			int i = Integer.parseInt(st.nextToken())-1;
			int l = Integer.parseInt(st.nextToken())-1;
			int k = Integer.parseInt(st.nextToken())-1;
			
			long ans = 0;
			for(int start = 0;start<11;start++)
			{
				int cur = count[start][k][l];
				if (i > 0) cur -= count[start][i - 1][l];
				if (j > 0) cur -= count[start][k][j - 1]; 
				if (i > 0 && j > 0) cur += count[start][i - 1][j - 1];
				
				ans += 1l*((start + t%c)%c) * cur;
			}
			sb.append(ans+"\n");
		}
		System.out.print(sb);
	}
	
}