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.lang.Math;

import java.util.*;

public class stt {
	
	
	public static void main(String[] args) {
    //int sz = 1 << 15;
 //   Scanner in = new Scanner(new BufferedInputStream(System.in, sz));
 //   PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out, sz));
		Scanner s= new Scanner(System.in);
        int n=s.nextInt();
        int views=s.nextInt();
        int c=s.nextInt();
        int data[][][]= new int[c+1][101][101];
        for(int i=0;i<n;i++)
        { int x= s.nextInt();
          int y= s.nextInt();
          int b= s.nextInt();
          data[b][x][y]++;
        	
        }
       for(int b=0;b<=c;b++) 
       {  for(int x=1;x<101;x++)
        	for(int y=2;y<101;y++)
        		data[b][x][y]+=data[b][x][y-1];
       for(int x=2;x<101;x++)
           for(int y=1;y<101;y++)
        	
        		data[b][x][y]+=data[b][x-1][y];
       }
       
       for(int view=1;view<=views;view++) 
       {  int time=s.nextInt();
    	   int x1=s.nextInt();
    	   int y1=s.nextInt();
    	   int x2=s.nextInt();
    	   int y2=s.nextInt();
    	   int sum=0;
    	   for(int b=0;b<=c;b++) 
           { int count = data[b][x2][y2];
             if(x1>=2)
             count-=data[b][x1-1][y2];
             if(y1>=2)
             count-=data[b][x2][y1-1];
             if(x1>=2 & y1>=2)
             count+=data[b][x1-1][y1-1];
             
            // System.out.println("view"+view+"b"+b+"="+count);
             
             sum+= count*((b+time)%(c+1));
             
             
    		   
            }
    	   System.out.println(sum);
    	   
       }
            
       
       
	} 
       
	}
