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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class ProblemA {

	static Scanner sc = new Scanner(System.in);
	static PrintWriter pw = new PrintWriter(System.out);
	

	public static void main(String[] args) throws IOException {
		int n = sc.nextInt() , q = sc.nextInt() , c = sc.nextInt() ;
		int count[][][] = new int[101][101][11] ;
		for(int i= 0 ; i < n ;i++){
			int xi = sc.nextInt()  , yi  = sc.nextInt() , si = sc.nextInt() ;
			count[xi][yi][si]++ ;
		}
		
		int preSum[][][] = new int[101][101][11] ;
		for(int i = 0 ; i < 11 ; i ++){
			for(int j = 1 ; j <101 ; j++){
				for(int k = 1; k< 101 ; k++){
					preSum[j][k][i] = preSum[j-1][k][i] +preSum[j][k-1][i] 
							-preSum[j-1][k-1][i] + count[j][k][i]+preSum[j][k][i];
				}
			}
		}
		for(int i = 0  ; i < q ; i ++){
			int t = sc.nextInt() , x1 = sc.nextInt() ,y1 = sc.nextInt() , x2 = sc.nextInt() , y2 = sc.nextInt() ;
			long ans = 0 ;
			for(int p = 0 ; p < 11 ; p++){
				long amount = preSum[x2][y2][p] - preSum[x1-1][y2][p] - preSum[x2][y1-1][p] + preSum[x1-1][y1-1][p];
				ans+= ((t+p)%(c+1))* amount ;
			}
			pw.println(ans);
		}
		pw.flush();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

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