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.*;

public class starsky
{
   public static void main(String[] args)
   {
      Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      int q = sc.nextInt();
      int c = sc.nextInt();
      //int[][] star = new int[n][3];
      int[][][] coor = new int[101][101][11];
      for (int k = 0; k < n; k++)
      {
         int x = sc.nextInt();
         int y = sc.nextInt();
         int s = sc.nextInt();
         for (int j = 0; j <= c; j++)
            coor[x][y][j]+=(j+s)%(c+1);
      }
      for (int k = 0; k <= c; k++)
      {
         for (int a = 1; a <= 100; a++)
         {
            for (int b = 1; b <= 100; b++)
            {
               coor[a][b][k] += coor[a-1][b][k]+coor[a][b-1][k]-coor[a-1][b-1][k];
            }
         }
      }
      
      for (int k = 0; k < q; k++)
      {
         int t = sc.nextInt();
         int x1 = sc.nextInt();
         int y1 = sc.nextInt();
         int x2 = sc.nextInt();
         int y2 = sc.nextInt();
         t%=(c+1);
         System.out.println(coor[x2][y2][t] - coor[x1-1][y2][t] - coor[x2][y1-1][t]+coor[x1-1][y1-1][t]);
      }
      /*
      for (int k = 0; k < n; k++)
      {
         star[k][0] = sc.nextInt();
         star[k][1] = sc.nextInt();
         star[k][2] = sc.nextInt();
      }
      for (int k = 0; k < q; k++)
      {
         int t = sc.nextInt();
         int x1 = sc.nextInt();
         int y1 = sc.nextInt();
         int x2 = sc.nextInt();
         int y2 = sc.nextInt();
         int sum = 0;
         for (int j = 0; j < n; j++)
         {
            if (x1 <= star[j][0] && star[j][0] <= x2 && y1 <= star[j][1] && star[j][1] <= y2)
               sum+=(star[j][2]+t)%(c+1);
         }
         System.out.println(sum);
      }*/
   }
}