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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class CF427C {

	public static int[][][] arr = new int[11][101][101]; // each max brightness,
															// each mod,

	public static int query(int cur, int x1, int x2, int y1, int y2) {
		return arr[cur][x2][y2] - arr[cur][x1 - 1][y2] - arr[cur][x2][y1 - 1] + arr[cur][x1 - 1][y1 - 1];
	}

	public static void main(String[] args) {
		MyScanner in = new MyScanner();
		int n = in.nextInt();
		int q = in.nextInt();
		int c = in.nextInt();

		for (int i = 0; i < n; i++) {
			int x = in.nextInt();
			int y = in.nextInt();
			int s = in.nextInt();
			for (int k = 0; k < c + 1; ++k) {
				arr[k][x][y] += ((s + k) % (c + 1));
			}
		}

		for (int k = 0; k < 11; k++) {
			for (int i = 1; i < 101; i++)
				for (int j = 1; j < 101; j++)
					arr[k][i][j] += arr[k][i][j - 1] + arr[k][i - 1][j] - arr[k][i - 1][j - 1];
		}

		for (int i = 0; i < q; ++i) {
			int t = in.nextInt(), x1 = in.nextInt(), y1 = in.nextInt(), x2 = in.nextInt(), y2 = in.nextInt();
			System.out.println(query(t % (c + 1), x1, x2, y1, y2));
		}
	}

	public static class MyScanner {
		BufferedReader br;
		StringTokenizer st;

		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

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

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

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}
