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 Stars {
    public static void main(String[] args) {
        SkyStars rv = new SkyStars();
        rv.input();
    }
}
class SkyStars{
    int n,q,c,x1,y1,x2,y2,s,t,i,j,k;
    int cnt[][][];

    Scanner scan = new Scanner(System.in);
    void input(){
        n = scan.nextInt();
        q = scan.nextInt();
        c = scan.nextInt();
        cnt = new int[c+1][101][101];

        for(i=0;i<n;i++){
            x1 = scan.nextInt();
            y1 = scan.nextInt();
            s = scan.nextInt();
            cnt[s][x1][y1]++;
        }

        compute();

        int res;
        for(i=0;i<q;i++){
            t = scan.nextInt();
            x1 = scan.nextInt();
            y1 = scan.nextInt();
            x2 = scan.nextInt();
            y2 = scan.nextInt();
            res = output(t,x1,y1,x2,y2);
            System.out.println(res);
        }
    }

    void compute(){
        for(k=0;k<=c;k++){
            for(i=1;i<101;i++){
                for(j=1;j<101;j++){
                    cnt[k][i][j] += cnt[k][i-1][j]+cnt[k][i][j-1]-cnt[k][i-1][j-1];
                }
            }
        }
    }

    int output(int t,int x1,int y1,int x2,int y2){
        int amount,bright=0,res=0;
        for(k=0;k<=c;k++){
            bright = (t+k)%(c+1);
            amount = cnt[k][x2][y2]-cnt[k][x1-1][y2]-cnt[k][x2][y1-1]+cnt[k][x1-1][y1-1];
            res += bright*amount;
        }
        return res;
    }
}
