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.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;

public class Q3 {

	static class FastScanner {
	    private InputStream stream;
	    private byte[] buf = new byte[1024];
	    private int curChar;
	    private int numChars;
	    private SpaceCharFilter filter;
	 
	    public FastScanner(InputStream stream) {
	        this.stream = stream;
	    }
	 
	    public static boolean isWhitespace(int c) {
	        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	    }
	 
	    public int read() {
	        if (numChars == -1)
	            throw new InputMismatchException();
	        if (curChar >= numChars) {
	            curChar = 0;
	            try {
	                numChars = stream.read(buf);
	            } catch (IOException e) {
	                throw new InputMismatchException();
	            }
	            if (numChars <= 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 {
	            if (c < '0' || c > '9')
	                throw new InputMismatchException();
	            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 {
	            if (c < '0' || c > '9')
	                throw new InputMismatchException();
	            res *= 10;
	            res += c - '0';
	            c = read();
	        } while (!isSpaceChar(c));
	        return res * sgn;
	    }
	 
	    public String nextToken() {
	        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 boolean isSpaceChar(int c) {
	        if (filter != null)
	            return filter.isSpaceChar(c);
	        return isWhitespace(c);
	    }
	 
	    public interface SpaceCharFilter {
	        public boolean isSpaceChar(int ch);
	    }
	}
	
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
	
	public static void main(String[] args) throws Exception {
		// TODO Auto-generated method stub
		FastScanner fs = new FastScanner(System.in);
		int n = fs.nextInt(), q = fs.nextInt(), c = fs.nextInt();
		int[][] arr = new int[n][3];
		for(int i=0; i<n; i++) {
			for(int j=0; j<3; j++) {
				arr[i][j] = fs.nextInt();
			}
		}
		
		int[][][] dp = new int[110][110][11];
		for(int i=0; i<n; i++)
			dp[arr[i][0]][arr[i][1]][arr[i][2]]++;
		
		long[][][] sum = new long[110][110][11];
		for(int i=1; i<110; i++)
			for(int j=1; j<110; j++)
				for(int k=0; k<=c; k++)
					sum[i][j][k] = (dp[i][j][k] + sum[i-1][j][k] + sum[i][j-1][k] - sum[i-1][j-1][k]);
		
		while(q-- > 0) {
			int t = fs.nextInt(), x1 = fs.nextInt(), y1 = fs.nextInt(), x2 = fs.nextInt(), y2 = fs.nextInt();
			long ans = 0;
			for(int i=0; i<=c; i++) {
				long temp = sum[x2][y2][i] - sum[x2][y1-1][i] - sum[x1-1][y2][i] + sum[x1-1][y1-1][i];
				ans += (temp * ((i + t)%(c+1)));
			}
			pw.println(ans);
		}
		pw.flush();
		pw.close();
	}
	
}