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.*;
/*
2 4 10
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
11 1 1 5 5
12 3 2 3 2

 */
public class C {
	
	public static void main(String[] args) throws IOException {
		
		FastScanner qwe = new FastScanner(System.in);
		
		int n = qwe.nextInt();
		int q = qwe.nextInt();
		int c = qwe.nextInt();
		
		
		int[][][] all = new int[11][101][101];
		
		ArrayList<Integer>[][] stars = new ArrayList[101][101];
		for (int i = 0; i < stars.length; i++) {
			for (int j = 0; j < stars.length; j++) {
				stars[i][j] = new ArrayList<Integer>();
			}
		}
		
		//for(int i =0; i < stars.length; i++) Arrays.fill(stars[i], -1);
		
		for(int i =0; i < n; i++){
			stars[qwe.nextInt()][qwe.nextInt()].add(qwe.nextInt());
		}
		
		
		
		for(int t =0; t <= 10; t++){
			for(int y =1; y <= 100; y++){
				for(int x = 1; x <= 100; x++){
					
					int here = 0;
					for(int s : stars[x][y]){
						here += (s+t)%(c+1);
					}

					
					all[t][x][y] = here;
					all[t][x][y] += all[t][x-1][y];
					all[t][x][y] += all[t][x][y-1];
					all[t][x][y] -= all[t][x-1][y-1];
					
				}
			}
			
		}
			
		while(q-->0){
			int t = qwe.nextInt()%(c+1);
			int x1 = qwe.nextInt();
			int y1 = qwe.nextInt();
			int x2 = qwe.nextInt();
			int y2 = qwe.nextInt();
			
			int ans = all[t][x2][y2]-all[t][x1-1][y2]-all[t][x2][y1-1]+all[t][x1-1][y1-1];
			
			System.out.println(ans);
		}
		
		
		qwe.close();
		
	}
	
	static class FastScanner
	{
		BufferedReader br;
		StringTokenizer st;
		public FastScanner(InputStream in)
		{
			br = new BufferedReader(new InputStreamReader(in));
			st = new StringTokenizer("");
		}
		public String next() throws IOException
		{
			while(!st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int nextInt() throws IOException
		{
			return Integer.parseInt(next());
		}
		public void close() throws IOException
		{
			br.close();
		}
	}
	

}

