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.*;

public class cf{

   public static void main(String[] args){
      Scanner f = new Scanner(System.in);
      int n = f.nextInt();
      int q = f.nextInt();
      int c = f.nextInt();
      c++;
      int[] sky = new int[c*10000];
      for(int i = 0; i < n; i++){
         int x = f.nextInt();
         int y = f.nextInt();
         int s = f.nextInt();
         x--;
         y--;
         for(int j = 0; j < c; j++){
            sky[j*10000+100*x+y] += (s+j)%c;
         }
      }
      for(int i = 0; i < c; i++){
         for(int k = 0; k < 10000; k++){
            int x = k/100;
            int y = k%100;
            if(x > 0)
               sky[i*10000+100*x+y] += sky[i*10000+100*(x-1)+y];
            if(y > 0)
               sky[i*10000+100*x+y] += sky[i*10000+100*x+y-1];
            if (x >0 && y >0)
               sky[i*10000+100*x+y] -= sky[i*10000+100*(x-1)+y-1]  ;
         }
      }   
      for(int i = 0; i< q; i++){
         int t = f.nextInt();
         int x1 = f.nextInt();
         int y1 = f.nextInt();
         int x2 = f.nextInt();
         int y2 = f.nextInt();
         x1 -= 1;
         y1 -= 1;
         x2 -= 1;
         y2 -= 1;
         t %= c;
         int out = sky[t*10000+100*x2+y2];
         if( x1 > 0)
            out -= sky[t*10000+100*(x1-1)+y2];
         if (y1 > 0)
            out -= sky[t*10000+100*x2+y1-1];
         if (x1 > 0 && y1 > 0)
            out += sky[t*10000+100*(x1-1)+y1-1];
         System.out.println(out);
      }
   }
}