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.
//package CF;

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class A {

	public static void main(String[] args) throws Exception {
		Scanner bf = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = bf.nextInt(), q = bf.nextInt(), c = bf.nextInt()+1;
		int[][][] sum = new int[c][100 + 1][100 + 1];
		for (int i = 0; i < n; i++) {
			int x = bf.nextInt(), y = bf.nextInt(), s = bf.nextInt();
			for (int j = 0; j < sum.length; j++) {
				sum[j][x][y] += (s+j)%c;
			}
		}
		for (int k = 0; k < sum.length; k++) {
			for (int i = 0; i < sum[k].length; i++) {
				for (int j = 0; j < sum[k][i].length; j++) {
					if (i - 1 >= 0) {
						sum[k][i][j] += sum[k][i - 1][j];
					}
					if (j - 1 >= 0) {
						sum[k][i][j] += sum[k][i][j - 1];
					}
					if (i - 1 >= 0 && j - 1 >= 0) {
						sum[k][i][j] -= sum[k][i - 1][j - 1];
					}
				}
			}
		}

		for (int i = 0; i < q; i++) {
			int t = bf.nextInt()%c, x1 = bf.nextInt(), y1 = bf.nextInt(), x2 = bf.nextInt(), y2 = bf.nextInt();
			int ans = sum[t][x2][y2];
			ans -= sum[t][x1 - 1][y2];	
			ans -= sum[t][x2][y1 - 1];
			ans += sum[t][x1 - 1][y1 - 1];
			out.println(ans);
		}
		out.flush();
		out.close();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader fileReader) {
			br = new BufferedReader(fileReader);
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public boolean ready() throws IOException {
			return br.ready();
		}
	}
}