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.Scanner;
import java.util.StringTokenizer;

public class Main {

	
	public static void main(String[] args) {
		// Write your code here
	Scanner s=new Scanner(System.in);
      
      int n=s.nextInt();
      int q=s.nextInt();
      int c=s.nextInt();
      long bright[][][]=new long[101][101][c+1];
      for(int i=0;i<n;i++)
      {
		int xi=s.nextInt();
        int yi=s.nextInt();
        int si=s.nextInt();
        bright[xi][yi][si]++;
      }
      int mod =(int)Math.pow(10,9)+7;
      for(int i=1;i<=100;i++)
      {
        for(int j=1;j<=100;j++)
        {
          for(int k=0;k<=c;k++)
          {
            bright[i][j][k]=(bright[i][j][k]+bright[i-1][j][k]+bright[i][j-1][k]-bright[i-1][j-1][k]+mod)%mod;
          }
        }
      }
      for(int i=0;i<q;i++)
      {
        int t=s.nextInt();
        int x1=s.nextInt();
        int y1=s.nextInt();
        int x2=s.nextInt();
        int y2=s.nextInt();
        long ans=0;
        for(int j=0;j<=c;j++)
        {
          int brightness=(t+j)%(c+1);
        long stars=(bright[x2][y2][j]-bright[x1-1][y2][j]-bright[x2][y1-1][j]+bright[x1-1][y1-1][j]+mod)%mod;
        ans=(ans+brightness*stars)%mod;
        }
        System.out.println(ans);
      }
	}

}