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 Main {

    public static void main(String[] args) {
        Scanner scn = new Scanner(System.in);

        int stars = scn.nextInt();
        int queries = scn.nextInt();
        int max_brightness = scn.nextInt();

        int perview = max_brightness + 1;
        int[][][] t = new int[11][101][101];

        // initialize the per view
        for(int i=0; i<11; i++){
            for(int j=0; j<101; j++){
                for(int k=0; k<101; k++) {
                    t[i][j][k] = 0;
                }
            }
        }

        // put brightness of star respect to its place
        int x, y, b;
        for(int i=0; i<stars; i++) {
            x = scn.nextInt();
            y = scn.nextInt();
            b = scn.nextInt();
            for(int j=0; j<=max_brightness; j++){
                t[j][x][y] += b;
                b = (b+1)%(max_brightness+1);
//                if(b<=max_brightness){
//                    t[j][x][y] = b;
//                    b++;
//                } else {
//                    t[j][x][y] = 0;
//                }
            }
        }

        // prefix sum of matrix
        for(int i=0; i<=max_brightness; i++){
            for(int j=1; j<101; j++){
                for(int k=1; k<101; k++) {
                    if(k!= 0){
                        t[i][j][k] += t[i][j][k-1] + t[i][j-1][k] - t[i][j-1][k-1];
                    }
                }
            }
        }

        // per queries brightness
        int time,x1,y1,x2,y2,ans;
        for(int i=0; i<queries; i++){
            time = scn.nextInt();
            time %= (max_brightness + 1);
            x1 = scn.nextInt();
            y1 = scn.nextInt();
            x2 = scn.nextInt();
            y2 = scn.nextInt();

            ans = t[time][x2][y2] - t[time][x1-1][y2] - t[time][x2][y1-1] + t[time][x1-1][y1-1];
            System.out.println(ans);
        }
    }
}