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.*;
import java.util.*;

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

	Main () {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		Scanner s = new Scanner(System.in);

		int n = fs.nextInt();
		int q = fs.nextInt();
		int c = fs.nextInt();

		num = new int[11][101][101];

		for(int i = 0; i < n; i++) {
			int x = fs.nextInt(), y = fs.nextInt();
			num[fs.nextInt()][x][y] += 1;
		}
		

		for(int k = 0; k < num.length; k++) {
			for(int i = 0; i < num[0].length; i++) {
				for(int j = 0; j < num[0].length; j++) {
					if(i > 0) num[k][i][j] += num[k][i-1][j];
					if(j > 0) num[k][i][j] += num[k][i][j-1];
					if(i > 0 && j > 0) num[k][i][j] -= num[k][i-1][j-1];
				}
			}
		}

//		for(int k = 0; k <= c; k++) {
//			for(int i = 0; i < 10; i++) {
//				for(int j = 0; j < 10; j++) {
//					System.out.printf("B = %d i = %d j = %d cnt = %d\n", k, i, j, num[k][i][j]);
//				}
//				System.out.println();
//			}
//			System.out.println();
//		}
		while (q-- > 0) {
			int time = fs.nextInt();
			int x1 = fs.nextInt(), y1 = fs.nextInt();
			int x2 = fs.nextInt(), y2 = fs.nextInt();
			long res = 0;
			for(int i = 0; i <= c; i++) {
//				System.out.printf("Brightness %d Size = %d\n", i, num[i][x2][y2] - num[i][x1-1][y1-1]);
				long add = (i + time) % (c + 1);
				add *= num[i][x2][y2] - num[i][x1-1][y2] - num[i][x2][y1-1] + num[i][x1-1][y1-1];
				res += add;
			}
//			System.out.println();
			out.println(res);
		}


		out.close();
	}

	int[][][] num;

	int[][] copy (int[][] a) {
		int[][] res = new int[a.length][a[0].length];
		for(int i = 0; i < a.length; i++) {
			for(int j = 0; j < a[0].length; j++) {
				res[i][j] = a[i][j];
			}
		}
		return res;
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		public FastScanner() {
			try	{
				br = new BufferedReader(new InputStreamReader(System.in));
				//				br = new BufferedReader(new FileReader("testdata.out"));
				st = new StringTokenizer("");
			} catch (Exception e){e.printStackTrace();}
		}
		public String next() {
			if (st.hasMoreTokens())	return st.nextToken();
			try {st = new StringTokenizer(br.readLine());}
			catch (Exception e) {e.printStackTrace();}
			return st.nextToken();
		}
		public int nextInt() {return Integer.parseInt(next());}
		public long nextLong() {return Long.parseLong(next());}
		public double nextDouble() {return Double.parseDouble(next());}
		public String nextLine() {
			String line = "";
			try {line = br.readLine();}
			catch (Exception e) {e.printStackTrace();}
			return line;
		}
		public Integer[] nextIntegerArray(int n) {
			Integer[] a = new Integer[n];
			for(int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}
		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for(int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}
		public char[] nextCharArray() {return nextLine().toCharArray();}
	}
}