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.*;
import java.io.*;

public class C835 {
	public static void main(String[] args) throws Exception {
		int n = i(), q = i(), c = i();

		int[][][] dp = new int[101][101][c + 1];

		for (int i = 0; i < n; i++) {
			dp[i()][i()][i()]++;
		}

		for (int x = 1; x < 101; x++) {
			for (int y = 1; y < 101; y++) {
				for (int z = 0; z <= c; z++) {
					dp[x][y][z] += dp[x - 1][y][z] + dp[x][y - 1][z] - dp[x - 1][y - 1][z];
				}
			}
		}

		for (int qq = 0; qq < q; qq++) {
			int t = i(), x1 = i(), y1 = i(), x2 = i(), y2 = i();

			int ans = 0;

			for (int i = 0; i <= c; i++) {
				ans += dp[x2][y2][i] * ((i + t) % (c + 1));
				ans -= dp[x1 - 1][y2][i] * ((i + t) % (c + 1));
				ans -= dp[x2][y1 - 1][i] * ((i + t) % (c + 1));
				ans += dp[x1 - 1][y1 - 1][i] * ((i + t) % (c + 1));
			}

			out.println(ans);
		}
		
		out.close();
	}

	static BufferedReader in;
	static StringTokenizer st = new StringTokenizer("");
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	static {
		try {
			in = new BufferedReader(new FileReader("cf.in"));
		} catch (Exception e) {
			in = new BufferedReader(new InputStreamReader(System.in));
		}
	}
	static void check() throws Exception {
		while (!st.hasMoreTokens()) st = new StringTokenizer(in.readLine());
	}
	static int i() throws Exception {check(); return Integer.parseInt(st.nextToken());}
	static String s() throws Exception {check(); return st.nextToken();}
	static double d() throws Exception {check(); return Double.parseDouble(st.nextToken());}
	static long l() throws Exception {check(); return Long.parseLong(st.nextToken());}
}
