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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author AEroui
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int q = in.nextInt();
            int c = in.nextInt();

            int[][][] mat = new int[110][110][15];
            int[][][] pre = new int[110][110][15];
            for (int i = 0; i < n; ++i) {
                int x = in.nextInt();
                int y = in.nextInt();

                int s = in.nextInt();

                for (int j = 0; j <= 10; ++j) {
                    mat[x][y][j] += ((s + j) % (c + 1));
                }
            }


            for (int i = 1; i <= 100; ++i) {
                int[] cur = new int[15];
                for (int j = 1; j <= 100; ++j) {
                    for (int k = 0; k <= 10; ++k) {
                        pre[i][j][k] = pre[i - 1][j][k] + mat[i][j][k] + cur[k];
                        cur[k] += mat[i][j][k];
                    }
                }
            }
            
            /*for (int i = 1; i <= 5; ++i) {
                for (int j = 1; j <= 5; ++j) {
                    out.print(pre[i][j][2] + " ");
                }
                out.println();
            }*/


            while (q-- > 0) {
                int t = in.nextInt();
                int x1 = in.nextInt();
                int y1 = in.nextInt();
                int x2 = in.nextInt();
                int y2 = in.nextInt();

                int p = pre[x2][y2][t % (c + 1)] - pre[x2][y1 - 1][t % (c + 1)] - pre[x1 - 1][y2][t % (c + 1)] + pre[x1 - 1][y1 - 1][t % (c + 1)];
                out.println(p);
            }
        }

    }
}

