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.InputStreamReader;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;

/**
 *
 * @author Fuad
 */
//1 4 5 6
public class Codeforces {

    static class pair<T1, T2> implements Comparator {

        T1 first;
        T2 second;

        pair(T1 f, T2 s) {
            first = f;
            second = s;
        }

        pair() {

        }

        @Override
        public int compare(Object o1, Object o2) {
            pair<Integer, Integer> p1 = (pair) o1;
            pair<Integer, Integer> p2 = (pair) o2;

            if (p1.first < p2.first) {
                return -1;
            } else if (p2.first < p1.first) {
                return 1;
            } else {
                return p1.second <= p2.second ? -1 : 1;
            }
        }

    }

    static int st[][][] = new int[11][101][101], rec[][][] = new int[11][101][101];

    public static void main(String[] args) throws Exception {
        // TODO code application logic here

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        String inp[];
        StringBuilder sol = new StringBuilder();
        int n, q, c;

        inp = br.readLine().split(" ");
        n = Integer.parseInt(inp[0]);
        q = Integer.parseInt(inp[1]);
        c = Integer.parseInt(inp[2]);

        for (int i = 0; i < n; i++) {
            int s, x, y;
            inp = br.readLine().split(" ");
            x = Integer.parseInt(inp[0]);
            y = Integer.parseInt(inp[1]);
            s = Integer.parseInt(inp[2]);
            st[s][x][y]++;

        }

        for (int i = 0; i <= c; i++) {
            for (int x = 1; x <= 100; x++) {
                for (int y = 1; y <= 100; y++) {
                    rec[i][x][y] = rec[i][x - 1][y] + rec[i][x][y - 1] - rec[i][x - 1][y - 1] + st[i][x][y];
                }
            }
        }

        for (int i = 0; i < q; i++) {

            int t, x1, y1, x2, y2;
            inp = br.readLine().split(" ");
            t = Integer.parseInt(inp[0]);
            x1 = Integer.parseInt(inp[1]);
            y1 = Integer.parseInt(inp[2]);
            x2 = Integer.parseInt(inp[3]);
            y2 = Integer.parseInt(inp[4]);

            long b = 0;
            for (int s = 0; s <= c; s++) {
                b += (((s + t) % (c + 1)) * (rec[s][x2][y2] - rec[s][x1 - 1][y2] - rec[s][x2][y1 - 1] + rec[s][x1 - 1][y1 - 1]));
            }
            sol.append(b).append("\n");
        }

        System.out.print(sol);
    }

    public static boolean isPalind(StringBuilder s) {

        int i = 0;
        int j = s.length() - 1;

        while (i < j) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
            i++;
            j--;
        }

        return true;
    }

}
