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.IOException;
import java.io.InputStream;
import java.util.NoSuchElementException;


public class Main {
  public static void main(String[] args) {
    FastScanner sc = new FastScanner();
    int n = sc.nextInt();
    int q = sc.nextInt();
    int c = sc.nextInt();
    int[][] star = sc.nextIntTable(n, 3); // x, y, s
    int[][] query = sc.nextIntTable(q, 5); // t, x1, y1, x2, y2
    
    
    int[][][] sum = new int[11][101][101];
    for (int[] st : star) {
      int x = st[0];
      int y = st[1];
      int s = st[2];
      
      sum[s][x][y] ++;
    }
    
    for (int s = 0; s <= 10; s ++) {
      for (int x = 1; x <= 100; x ++) {
        for (int y = 1; y <= 100; y ++) {
          sum[s][x][y] += sum[s][x][y - 1];
        }

        for (int y = 1; y <= 100; y ++) {
          sum[s][x][y] += sum[s][x - 1][y];
        }
      }
    }
    
    
    StringBuilder sb = new StringBuilder();
    for (int[] qr : query) {
      int t = qr[0];
      int x1 = qr[1];
      int y1 = qr[2];
      int x2 = qr[3];
      int y2 = qr[4];
      
      long now = 0;
      for (int s = 0; s <= 10; s ++) {
        int v = sum[s][x2][y2] - sum[s][x1 - 1][y2] - sum[s][x2][y1 - 1] + sum[s][x1 - 1][y1 - 1];
        
        now += v * ((t + s) % (c + 1));
      }
      sb.append(now);
      sb.append("\n");
    }
    System.out.println(sb);
  }

}


class FastScanner {
	public static String debug = null;

	private final InputStream in = System.in;
	private int ptr = 0;
	private int buflen = 0;
	private byte[] buffer = new byte[1024];
	private boolean eos = false;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				if (debug != null) {
					buflen = debug.length();
					buffer = debug.getBytes();
					debug = "";
					eos = true;
				} else {
					buflen = in.read(buffer);
				}
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen < 0) {
				eos = true;
				return false;
			} else if (buflen == 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	private void skipUnprintable() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
	}

	public boolean isEOS() {
		return this.eos;
	}

	public boolean hasNext() {
		skipUnprintable();
		return hasNextByte();
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}

	public int nextInt() {
		return (int) nextLong();
	}

	public long[] nextLongList(int n) {
		return nextLongTable(1, n)[0];
	}

	public int[] nextIntList(int n) {
		return nextIntTable(1, n)[0];
	}

	public long[][] nextLongTable(int n, int m) {
		long[][] ret = new long[n][m];
		for (int i = 0; i < n; i ++) {
			for (int j = 0; j < m; j ++) {
				ret[i][j] = nextLong();
			}
		}
		return ret;
	}

	public int[][] nextIntTable(int n, int m) {
		int[][] ret = new int[n][m];
		for (int i = 0; i < n; i ++) {
			for (int j = 0; j < m; j ++) {
				ret[i][j] = nextInt();
			}
		}
		return ret;
	}
}