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.awt.geom.*;
import java.io.*;
import java.math.*;
import java.text.*; 
import java.util.*;
import java.util.regex.*;

/*
		  br = new BufferedReader(new FileReader("input.txt"));
		  pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
		  br = new BufferedReader(new InputStreamReader(System.in));
		  pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
 */

public class Main {
	private static BufferedReader br;
	private static StringTokenizer st;
	private static PrintWriter pw;

	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		//int qq = 1;
		int qq = Integer.MAX_VALUE;
		//int qq = readInt();
		for(int casenum = 1; casenum <= qq; casenum++)	{
			int n = readInt();
			int qqq = readInt();
			int k = readInt();
			int[][][] dp = new int[k+1][101][101];
			while(n-- > 0) {
				int x = readInt();
				int y = readInt();
				int start = readInt();
				for(int i = 0; i <= k; i++) {
					dp[i][x][y] += (i+start)%(k+1);
				}
			}
			for(int[][] internal: dp) {
				for(int i = 1; i <= 100; i++) {
					for(int j = 1; j <= 100; j++) {
						internal[i][j] += internal[i][j-1] + internal[i-1][j] - internal[i-1][j-1];
					}
				}
			}
			while(qqq-- > 0) {
				int t = readInt()%(k+1);
				int x1 = readInt();
				int y1 = readInt();
				int x2 = readInt();
				int y2 = readInt();
				pw.println(dp[t][x2][y2] - dp[t][x1-1][y2] - dp[t][x2][y1-1] + dp[t][x1-1][y1-1]);
			}
		}
		exitImmediately();
	}

	private static void exitImmediately() {
		pw.close();
		System.exit(0);
	}

	private static long readLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	private static double readDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	private static int readInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	private static String nextLine() throws IOException  {
		String s = br.readLine();
		if(s == null) {
			exitImmediately();
		}
		st = null;
		return s;
	}

	private static String nextToken() throws IOException  {
		while(st == null || !st.hasMoreTokens())  {
			st = new StringTokenizer(nextLine().trim());
		}
		return st.nextToken();
	}
}