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.Arrays;
import java.util.Scanner;

public class Main {

    private void run(){
        Scanner in = new Scanner(System.in);
        int[][][] data = new int[100][100][11];

        int n = in.nextInt();
        int q = in.nextInt();
        int c = in.nextInt();
        int sum, x1, x2, y1, y2, t, s;

        for(int i = 0; i < n; ++i){
            data[in.nextInt()-1][in.nextInt()-1][in.nextInt()]++;
        }

        for(int p = 0; p < 11; ++p){
            for(int i = 1; i < 100; ++i){
                data[i][0][p] = data[i - 1][0][p] + data[i][0][p];
                data[0][i][p] = data[0][i - 1][p] + data[0][i][p];
            }
        }

        for(int p = 0; p < 11; ++p){
            for(int i = 1; i < 100; ++i){
                for(int j = 1; j < 100; ++j){
                    data[i][j][p] = data[i - 1][j][p] + data[i][j - 1][p] - data[i - 1][j - 1][p] + data[i][j][p];
                }
            }
        }

        for (int i = 0; i < q; ++i){
            int amount;
            t = in.nextInt();
            x1 = in.nextInt() - 1;
            y1 = in.nextInt() - 1;
            x2 = in.nextInt() - 1;
            y2 = in.nextInt() - 1;
            sum = 0;
            for(int p = 0; p < 11; ++p){
                amount = data[x2][y2][p];

                if (x1 > 0 || y1 > 0){
                    if (x1 > 0){ amount -= data[x1 - 1][y2][p]; }
                    if (y1 > 0) { amount -= data[x2][y1 - 1][p]; }
                    if (x1 > 0 && y1 > 0) { amount += data[x1 - 1][y1 - 1][p]; }
                }

                sum += ((p + t) % (c + 1)) * amount;
            }
            System.out.println(sum);
        }
    }

    public static void main(String[] args) { new Main().run(); }
}