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.IOException;
import java.io.PrintWriter;
import java.net.Inet4Address;
import java.util.*;
import java.io.*;

public class Main {
    public void solve() throws IOException {
        int n = nextInt();
        int q = nextInt();
        int c = nextInt();
        int[][][] a = new int[101][101][c + 1];
        for (int i = 0; i < n; i++) {
            int x = nextInt();
            int y = nextInt();
            int z = nextInt();
            for (int j = 0; j <= c; j++) {
                a[x][y][j] += (z + j) % (c + 1);
            }
        }
        for (int i = 0; i < 101; i++) {
            for (int j = 0; j < 101; j++) {
                for (int k = 0; k <= c; k++) {
                    if (i > 0 && j > 0) {
                        a[i][j][k] += a[i - 1][j][k] + a[i][j - 1][k] - a[i - 1][j - 1][k];
                    } else {
                        if (i > 0) {
                            a[i][j][k] += a[i - 1][j][k];
                        }
                        if (j > 0) {
                            a[i][j][k] += a[i][j - 1][k];
                        }
                    }
                }
            }
        }
        for (int ttt = 0; ttt < q; ttt++) {
            int t = nextInt() % (c + 1);
            int x1 = nextInt();
            int y1 = nextInt();
            int x2 = nextInt();
            int y2 = nextInt();
            int z = a[x2][y2][t] + a[x1 - 1][y1 - 1][t] - a[x1 - 1][y2][t] - a[x2][y1 - 1][t];
            out.println(z);
        }
    }


    BufferedReader br;
    StringTokenizer sc;
    PrintWriter out;

    public String nextToken() throws IOException {
        while (sc == null || !sc.hasMoreTokens()) {
            try {
                sc = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                return null;
            }
        }
        return sc.nextToken();
    }

    public boolean hasNext() throws IOException {
        while (sc == null || !sc.hasMoreTokens()) {
            try {
                sc = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                return false;
            }
        }
        return true;
    }

    public Integer nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public Long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public Double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public static void main(String[] args) throws IOException {
        try {
            Locale.setDefault(Locale.US);
        } catch (Exception e) {
        }
        new Main().run();
    }

    public void run() {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
//            br = new BufferedReader(new FileReader("a"));
//            out = new PrintWriter(new File("a"));
            solve();
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }
}