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

public class C {

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);

        Solver solver = new Solver();
        solver.solve(1, in, out);

        out.close();
    }

    static class Solver {

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int q = in.nextInt();
            int c = in.nextInt();

            int[][][] rect = new int[100][100][11];
            int[][][] map = new int[100][100][11];
            for (int i = 0; i < n; i++) {
                int x = in.nextInt() - 1;
                int y = in.nextInt() - 1;
                int s = in.nextInt();
                map[x][y][s] += 1;
            }
            for (int k = 0; k <= 10; k++) {
                rect[0][0][k] = map[0][0][k];
            }
            for (int i = 1; i < 100; i++) {
                for (int k = 0; k <= 10; k++) {
                    rect[0][i][k] = rect[0][i - 1][k] + map[0][i][k];
                    rect[i][0][k] = rect[i - 1][0][k] + map[i][0][k];
                }
            }
            for (int i = 1; i < 100; i++) {
                for (int j = 1; j < 100; j++) {
                    for (int k = 0; k <= 10; k++) {
                        rect[i][j][k] = rect[i - 1][j][k] + rect[i][j - 1][k] - rect[i - 1][j - 1][k] + map[i][j][k];
                    }
                }
            }
            for (int query = 0; query < q; query++) {
                int t = in.nextInt();
                int x1 = in.nextInt() - 1;
                int y1 = in.nextInt() - 1;
                int x2 = in.nextInt() - 1;
                int y2 = in.nextInt() - 1;
                int[] b = new int[11];
                for (int k = 0; k <= 10; k++) {
                    b[k] = rect[x2][y2][k];
                    if (x1 > 0) b[k] -= rect[x1 - 1][y2][k];
                    if (y1 > 0) b[k] -= rect[x2][y1 - 1][k];
                    if (x1 > 0 && y1 > 0) b[k] += rect[x1 - 1][y1 - 1][k];
                }
                long res = 0;
                for (int k = 0; k <= 10; k++) {
                    long next = (1L * (k + t)) % (c + 1);
                    res += b[k] * next;
                }
                out.println(res);
            }
        }

    }

    static class InputReader {

        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public float nextFloat() {
            return Float.parseFloat(next());
        }

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

    }
}
