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.text.*;
    import java.math.*;
    import java.util.regex.*;
    import java.lang.*;
    import java.io.*;
        
    public class Solution {
     
    public static void main( String args[] ){
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int q = in.nextInt();
        int c = in.nextInt();
        
        int accumulator [][][] = new int [102][102][c+1];
        for (int i = 0; i < n; i++) {
            int x = in.nextInt();
            int y = in.nextInt();
            int s = in.nextInt();
            for (int j = 0; j < c+1; j++) {
                accumulator[x][y][j] += (s+j)%(c+1);
            }
        }
        
        /*
        for (int l = 0; l < c+1; l++) {
            System.out.println("level accumulator "+ l);
            for (int i = 0; i < 5; i++) {
                for (int j = 0; j < 5; j++) {
                    System.out.print(accumulator[i][j][l]+" ");
                }
                System.out.println();
            }
        }*/
        
        int accumulator_new [][][] = new int [102][102][c+1];
        for (int l = 0; l < c+1; l++) {
            for (int i = 1; i < 102; i++) {
                for (int j = 1; j < 102; j++) {
                    //accumulator[i][j][l] = accumulator[i][j][l] + accumulator[i-1][j][l] + accumulator[i][j-1][l]; //+ accumulator[i-1][j-1][l];
                   /*int sum = 0;
                   for (int p = 1; p < i; p++) {
                       for (int r = 1; r < j; r++) {
                           sum += accumulator[p][q][l];
                       }
                   }
                    accumulator_new[i][j][l] = accumulator[i][j][l] + accumulator[i-1][j-1][l] + sum;
                    */
                    //dp[r + 1][c + 1] = dp[r + 1][c] + dp[r][c + 1] + matrix[r][c] - dp[r][c];
                    accumulator_new[i][j][l] = accumulator_new[i-1][j][l] + accumulator_new[i][j-1][l] + accumulator[i-1][j-1][l] - accumulator_new[i-1][j-1][l];
                }
            }
        }
        
        /*
        for (int l = 0; l < c+1; l++) {
            System.out.println("level accumulator_new "+ l);
            for (int i = 0; i < 5; i++) {
                for (int j = 0; j < 5; j++) {
                    System.out.print(accumulator_new[i][j][l]+" ");
                }
                System.out.println();
            }
        }*/
        
        for (int i = 0; i < q; i++) {
            int t = in.nextInt();
            int x1 = in.nextInt();
            int y1 = in.nextInt();
            int x2 = in.nextInt();
            int y2 = in.nextInt();
            
            int level = t%(c+1);
            //dp[row2 + 1][col2 + 1] - dp[row1][col2 + 1] - dp[row2 + 1][col1] + dp[row1][col1];\       
            int ans = accumulator_new[x2+1][y2+1][level] - accumulator_new[x1][y2+1][level] - accumulator_new[x2+1][y1][level] + accumulator_new[x1][y1][level];
            System.out.println(ans);
            
        }
        
        }
    }
