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.*;

public class C {

    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n = sc.nextInt(), q = sc.nextInt(), c = sc.nextInt();
        int cumCol[][][] = new int[c + 1][101][101];
        int[][] grid = new int[101][101];
        int[][][] exist = new int[101][101][11];
        while (n-- > 0) {
            int x, y, col;
            grid[x = (sc.nextInt() - 1)][y = (sc.nextInt() - 1)] = col = sc.nextInt();
            exist[x][y][col] ++;
        }

        for (int row = 0; row <= 100; row++) {
            for (int color = 0; color <= c; color++)
                cumCol[color][row][0] = exist[row][0][color];
            for (int col = 1; col <= 100; col++) {
                for (int color = 0; color <= c; color++)
                    cumCol[color][row][col] = exist[row][col][color];
                for (int color = 0; color <= c; color++)
                    cumCol[color][row][col] += cumCol[color][row][col - 1];
            }
        }

        while (q-- > 0) {
            int t = sc.nextInt(), row1 = sc.nextInt() - 1, col1 = sc.nextInt() - 1, row2 = sc.nextInt() - 1, col2 = sc.nextInt() - 1;
            long[] res = new long[c + 1];
            for (int row = Math.min(row1, row2); row <= Math.max(row1, row2); row++) {
                for (int color = 0; color <= c; color++) {
                    res[color] += cumCol[color][row][Math.max(col1, col2)] -
                            (Math.min(col1, col2) == 0 ? 0 : cumCol[color][row][Math.min(col1, col2) - 1]);
                }
            }
            long ret = 0;
            for (int i = 0; i <= c; i++) {
                ret += 1L * res[i] * ((i + t) % (c + 1));
            }
            out.println(ret);
        }
        out.flush();
        out.close();
    }

























    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();
        }

    }
}