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.Scanner;

public class WorkFile {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt(), q = s.nextInt(), c = s.nextInt();
        int[][][] matr = new int[109][109][c+1];
        for (int i=0; i<n; i++) {
            int x = s.nextInt(), y = s.nextInt(), ci = s.nextInt();
            matr[x][y][ci]++;
        }
        for (int i=0; i<109; i++) {
            for (int j=0; j<109; j++) {
                for (int k=0; k<=c; k++) {
                    if (j>0) {
                        matr[i][j][k]+=matr[i][j-1][k];
                        if (i>0) matr[i][j-1][k]+=matr[i-1][j-1][k];
                    }
                }
            }
            if (i>0) {
                for (int k=0; k<=c; k++) {
                    matr[i][108][k]+=matr[i-1][108][k];
                }
            }
        }
        StringBuilder ans = new StringBuilder();
        while (q-->0) {
            long t = s.nextInt();
            int x1 = s.nextInt(), y1 = s.nextInt();
            int x2 = s.nextInt(), y2 = s.nextInt();
            long result = 0;
            for (int i=0; i<=c; i++) {
                long elem = 0;
                if (t+i<=c) elem = t+i;
                else elem = (t-(c-i)-1)%(c+1);
                result += elem*(matr[x2][y2][i]-matr[x1-1][y2][i]-
                        matr[x2][y1-1][i]+matr[x1-1][y1-1][i]);
            }
            ans.append(result).append("\n");
        }
        System.out.println(ans);
    }
}