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.BufferedWriter;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.util.Scanner;
import java.util.ArrayList;

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

    private final int MAX_CORD = 101;

    private Scanner in;
    private BufferedWriter out;
    private int n, q, c;
    private int dp[][][];
    
    private int get(int t, int x1, int y1, int x2, int y2) {
        t %= c;
        return dp[t][x2][y2] - dp[t][x1 - 1][y2] - dp[t][x2][y1 - 1] + dp[t][x1 - 1][y1 - 1];
    }

    private void run() throws IOException {
        in = new Scanner(System.in);
        out = new BufferedWriter(new OutputStreamWriter(System.out));
        n = in.nextInt();
        q = in.nextInt();
        c = in.nextInt();
        c++;
        
        ArrayList[][] points = new ArrayList[MAX_CORD][MAX_CORD];
        for (int i = 0; i < MAX_CORD; ++i)
            for (int j = 0; j < MAX_CORD; ++j)
                points[i][j] = new ArrayList();

        for (int i = 0; i < n; i++) {
            int x = in.nextInt();
            int y = in.nextInt();
            int s = in.nextInt();
            points[x][y].add(s);
        }
        
        dp = new int[c][MAX_CORD][MAX_CORD];
        for (int r = 0; r < c; ++r) {
            for (int i = 1; i < MAX_CORD; ++i)
                for (int j = 1; j < MAX_CORD; ++j) {
                    int sum = 0;
                    for (int k = 0; k < points[i][j].size(); ++k)
                        sum += (int)points[i][j].get(k);

                    dp[r][i][j] = sum + dp[r][i - 1][j] + dp[r][i][j - 1] - dp[r][i - 1][j - 1];

                    for (int k = 0; k < points[i][j].size(); ++k) {
                        points[i][j].set(k, ((int)points[i][j].get(k)) + 1);
                        if (((int)points[i][j].get(k)) == c) {
                            points[i][j].set(k, 0);
                        }
                    }
                }
        }

        for (int i = 0; i < q; i++) {
            int t = in.nextInt();
            int x1 = in.nextInt();
            int y1 = in.nextInt();
            int x2 = in.nextInt();
            int y2 = in.nextInt();
            out.write(get(t, x1, y1, x2, y2) + "\n");
        }

        out.flush();
    }
}
