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.*;
import java.util.*;
import java.math.*;
import java.math.BigInteger.*;

public class C427
{
	public static void main(String[] args) throws Exception
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer s = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(s.nextToken());
		int q = Integer.parseInt(s.nextToken());
		int c = Integer.parseInt(s.nextToken());
		
		ArrayList[][] stars = new ArrayList[101][101];
		for (int i=1; i<=100; ++i)
			for (int j=1; j<=100; ++j)
				stars[i][j] = new ArrayList<Integer>();
		
		for (int i=0; i<n; ++i)
		{
			s = new StringTokenizer(br.readLine());
			int x1 = Integer.parseInt(s.nextToken());
			int x2 = Integer.parseInt(s.nextToken());
			int si = Integer.parseInt(s.nextToken());
			stars[x1][x2].add(si);
		}
		
		int[][][] doubleprefix = new int[c+1][101][101];
		for (int i=0; i<=c; ++i)
			for (int j=1; j<=100; ++j)
				for (int k=1; k<=100; ++k)
				{
					doubleprefix[i][j][k] = doubleprefix[i][j][k-1];
					for (int l=1; l<=j; ++l)
						for (Object o : stars[l][k])
							doubleprefix[i][j][k] += ((int)o+i)%(c+1);
				}
		
		for (int i=0; i<q; ++i)
		{
			s = new StringTokenizer(br.readLine());
			int t = Integer.parseInt(s.nextToken()) % (c+1);
			int x1 = Integer.parseInt(s.nextToken());
			int y1 = Integer.parseInt(s.nextToken());
			int x2 = Integer.parseInt(s.nextToken());
			int y2 = Integer.parseInt(s.nextToken());
			System.out.println(doubleprefix[t][x2][y2] - doubleprefix[t][x1-1][y2] - doubleprefix[t][x2][y1-1] + doubleprefix[t][x1-1][y1-1]);
		}
	}
}
