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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.Iterator;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.util.TreeSet;
import java.io.BufferedReader;
import java.io.InputStreamReader;

/**
 * Built using CHelper plug-in Actual solution is at the top
 */
public class Main {
	public final static double MAX_T = 1000000;
	public final static double MIN_T = 0;
	public static int s = 0;

	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskC solver = new TaskC();
		solver.solve(1, in, out);
		out.close();
	}

	static class TaskC {
		public void solve(int testNumber, InputReader in, PrintWriter out) {
			int n = in.nextInt();
			int q = in.nextInt();
			char c = (char) in.nextInt();
			c++;

			char[] x = new char[n];
			char[] y = new char[n];
			char[] s = new char[n];

			for (int i = 0; i < n; ++i) {
				x[i] = (char) in.nextInt();
				y[i] = (char) in.nextInt();
				s[i] = (char) in.nextInt();
			}

			char[] view_t = new char[q];
			char[] view_x1 = new char[q];
			char[] view_y1 = new char[q];
			char[] view_x2 = new char[q];
			char[] view_y2 = new char[q];

			for (int i = 0; i < q; ++i) {
				view_t[i] = (char) (in.nextInt() % c);
				view_x1[i] = (char) in.nextInt();
				view_x1[i]--;
				view_y1[i] = (char) in.nextInt();
				view_y1[i]--;
				view_x2[i] = (char) in.nextInt();
				view_y2[i] = (char) in.nextInt();
			}

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

			for (int i = 0; i < n; i++) {
				for (int t = 0; t < c; t++) {
					dp[x[i]][y[i]][t] += (s[i] + t) % c;
				}
			}

			for (int t = 0; t < c; t++) {
				for (int i = 0; i <= 100; i++) {
					int counter = 0;
					for (int j = 0; j <= 100; j++) {
						counter += dp[i][j][t];
						dp[i][j][t] = counter;
					}
				}
			}

			for (int t = 0; t < c; t++) {
				for (int j = 0; j <= 100; j++) {
					int counter = 0;
					for (int i = 0; i <= 100; i++) {
						counter += dp[i][j][t];
						dp[i][j][t] = counter;
					}
				}
			}

			for (int q1 = 0; q1 < q; q1++) {
				int brightness = 0;
				brightness += dp[view_x2[q1]][view_y2[q1]][view_t[q1]];
				brightness -= dp[view_x2[q1]][view_y1[q1]][view_t[q1]];
				brightness -= dp[view_x1[q1]][view_y2[q1]][view_t[q1]];
				brightness += dp[view_x1[q1]][view_y1[q1]][view_t[q1]];

				out.println(brightness);
			}

		}

	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

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

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

		public double nextFloat() {
			return Double.parseDouble(next());
		}

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

	}
}
