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;

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/**
 *
 * @author bbarbakadze
 */
public class CF_US_427C_Star_sky {
    public static void main(String[] Args) {
        Scanner in = new Scanner(System.in);
        int N = in.nextInt(), Q = in.nextInt(), C = in.nextInt();
        int[][][] M = new int[101][101][11];
        
        for (int i = 0; i < N; i++) {
            int x = in.nextInt(), y = in.nextInt(), c = in.nextInt();
            int cur = c;
            for (int t = 0; t <= 10; t++) {
                if (cur > C) cur = 0;
                M[x][y][t] += cur;
                cur++;
            }
        }
        
        int [][][] S = new int[101][101][11];
        for (int k = 0; k <= 10; k++) 
            for (int i = 1; i <= 100; i++)
                for (int j = 1; j <= 100; j++)
                    S[i][j][k] = S[i - 1][j][k] + S[i][j - 1][k] - S[i - 1][j - 1][k] + M[i][j][k];
        
        for (int i = 0; i < Q; i++) {
            int t = in.nextInt(), x1 = in.nextInt(), y1 = in.nextInt(), x2 = in.nextInt(), y2 = in.nextInt();
            t = t % (C + 1);
            int sum = S[x2][y2][t] - S[x2][y1 - 1][t] - S[x1 - 1][y2][t] + S[x1 - 1][y1 - 1][t];
            System.out.println(sum);
            
        }
        
    }
}
