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 DumbC {
	public DumbC() {
		Scanner s = new Scanner();

		int n = s.nextInt();
		int q = s.nextInt();
		int c = s.nextInt();
		StringBuilder sb = new StringBuilder();
		
		int[][][] pointCounts = new int[c+1][100][100];
		
//		HashMap<Rectangle, Integer>[] ds = new HashMap[c+1];
		
		for (int i = 0; i < n; i++){
			int x = s.nextInt() - 1;
			int y = s.nextInt() - 1;
			int si = s.nextInt();
			for (int f = 0; f < c+1; f++) {
				int bright = (si + f) % (c+1);
				pointCounts[f][x][y] += bright;
			}
		}
		
		
		for (int iq = 0; iq < q; iq++) {
			int t = s.nextInt() % (c+1);
			int x1 = s.nextInt()-1;
			int y1 = s.nextInt()-1;
			int x2 = s.nextInt()-1;
			int y2 = s.nextInt()-1;
			
			
			long total = 0;
		
			int[][] totals = pointCounts[t];
			for (int i = x1; i <= x2; i++) {
				for (int j = y1; j <= y2; j++) {
					total += totals[i][j];
				}
			}
			sb.append(total + "\n");
		
		}
		//sub 1!
		System.out.print(sb);
	}
	
	
	public class Point {
		public Point(int x, int y, int s) {
			super();
			this.x = x;
			this.y = y;
			this.s = s;
		}

		int x,y,s;
		
	}
	
	
	
	
	public class Rectangle {
		@Override
		public int hashCode() {
			final int prime = 31;
			int result = 1;
			result = prime * result + getOuterType().hashCode();
			result = prime * result + x1;
			result = prime * result + x2;
			result = prime * result + y1;
			result = prime * result + y2;
			return result;
		}

		@Override
		public boolean equals(Object obj) {
			if (this == obj)
				return true;
			if (obj == null)
				return false;
			if (getClass() != obj.getClass())
				return false;
			Rectangle other = (Rectangle) obj;
			if (!getOuterType().equals(other.getOuterType()))
				return false;
			if (x1 != other.x1)
				return false;
			if (x2 != other.x2)
				return false;
			if (y1 != other.y1)
				return false;
			if (y2 != other.y2)
				return false;
			return true;
		}

		@Override
		public String toString() {
			return "Rectangle [x1=" + x1 + ", y1=" + y1 + ", x2=" + x2
					+ ", y2=" + y2 + "]";
		}

		public Rectangle(int x1, int y1, int x2, int y2) {
			super();
			this.x1 = x1;
			this.y1 = y1;
			this.x2 = x2;
			this.y2 = y2;
		}

		int x1,y1,x2,y2;

		private DumbC getOuterType() {
			return DumbC.this;
		}
		
		
	}
	

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

	public static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		public Scanner(Reader in) {
			br = new BufferedReader(in);
		}

		public Scanner() {
			this(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 readNextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

}