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 {
	
	static InputReader in = new InputReader(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	
	
	public static void main(String[] args) {
		
		int n = in.nextInt();
		int q = in.nextInt();
		int c = in.nextInt();
		
		int[] s = new int[n];
		int[] x = new int[n];
		int[] y = new int[n];
		
		for(int i=0; i<n; ++i) {
			x[i] = in.nextInt();
			y[i] = in.nextInt();
			s[i] = in.nextInt();
		}
		
		
		int[][][] sum = new int[c+1][101][101];
		for(int i=0; i <= c; ++i) {
			int[][] b = new int[101][101];
			for(int j=0; j<n; ++j) {
				int xx = x[j];
				int yy = y[j];
				int ss = (s[j] + i) % (c+1);
				
				b[yy][xx] += ss;
			}
			
			for(int yi = 1; yi <= 100; ++yi) {
				for(int xi = 1; xi <= 100; ++xi) {
					sum[i][yi][xi] = sum[i][yi][xi-1] + sum[i][yi-1][xi] + b[yi][xi] - sum[i][yi-1][xi-1];
				}
			}
		}
		
//		System.out.println();
//		for(int i=0; i < 10; ++i) {
//			for(int j=0; j < 10; ++j) {
//				System.out.print( sum[2][i][j] + " " );
//			}
//			System.out.println();
//		}
		
		
		for(int i=0; i < q; ++i) {
			int t = in.nextInt();
			int x1 = in.nextInt();
			int y1 = in.nextInt();
			int x2 = in.nextInt();
			int y2 = in.nextInt();
			
			int mod = c+1;
			t %= mod;
			
			int ans = sum[t][y2][x2] - sum[t][y1-1][x2] - sum[t][y2][x1-1] + sum[t][y1-1][x1-1];
			System.out.println(ans);
		}
		
		out.close();
	}
	
	static void shuffle(int[] a) {
		Random r = new Random();
		for(int i = a.length - 1; i > 0; --i) {
			int si = r.nextInt(i);
			int t = a[si];
			a[si] = a[i];
			a[i] = t;
		}
	}
}


class InputReader {

	private final InputStream stream;
	private final byte[] buf = new byte[8192];
	private int curChar, snumChars;

	public InputReader(InputStream st) {
		this.stream = st;
	}

	public int read() {
		if (snumChars == -1)
			throw new InputMismatchException();
		if (curChar >= snumChars) {
			curChar = 0;
			try {
				snumChars = stream.read(buf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (snumChars <= 0)
				return -1;
		}
		return buf[curChar++];
	}

	public int nextInt() {
		int c = read();
		while (isSpaceChar(c)) {
			c = read();
		}
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		int res = 0;
		do {
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}

	public long nextLong() {
		int c = read();
		while (isSpaceChar(c)) {
			c = read();
		}
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		long res = 0;
		do {
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}

	public int[] nextIntArray(int n) {
		int a[] = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextInt();
		}
		return a;
	}

	public String readString() {
		int c = read();
		while (isSpaceChar(c)) {
			c = read();
		}
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isSpaceChar(c));
		return res.toString();
	}

	public String nextLine() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isEndOfLine(c));
		return res.toString();
	}

	public boolean isSpaceChar(int c) {
		return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	}

	private boolean isEndOfLine(int c) {
		return c == '\n' || c == '\r' || c == -1;
	}

}