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.*;
import java.util.*;

public class Main {
    private final int MAX = 100;
    private int[][][] count;
    private int n, q, c;
    public static void main(String[] args) throws IOException {
        new Main().run();
    }

    public void run() throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));

        StringTokenizer st = new StringTokenizer(br.readLine());
        n = Integer.parseInt(st.nextToken());
        q = Integer.parseInt(st.nextToken());
        c = Integer.parseInt(st.nextToken());
        count = new int[c + 1][MAX + 1][MAX + 1];

        for(int i = 0; i < n; i++) {
            st = new StringTokenizer(br.readLine());
            int x1 = Integer.parseInt(st.nextToken());
            int y1 = Integer.parseInt(st.nextToken());
            int s1 = Integer.parseInt(st.nextToken());
            count[s1][x1][y1]++;
        }

        for(int p = 0; p <= c; p++) {
            for(int i = 1; i <= MAX; i++) {
                for(int j = 1; j <= MAX; j++) {
                    count[p][i][j] += count[p][i-1][j] + count[p][i][j-1] - count[p][i-1][j-1];
                }
            }
        }

        for(int i = 0; i < q; i++) {
            st = new StringTokenizer(br.readLine());
            int vt = Integer.parseInt(st.nextToken());
            int vx1 = Integer.parseInt(st.nextToken());
            int vy1 = Integer.parseInt(st.nextToken());
            int vx2 = Integer.parseInt(st.nextToken());
            int vy2 = Integer.parseInt(st.nextToken());
            bw.write(solution(vt, vx1, vy1, vx2, vy2) + "\n");
        }

        bw.flush();
        bw.close();
    }

    private int solution(int vt, int vx1, int vy1, int vx2, int vy2) {
        int result = 0;
        for(int i = 0; i <= c; i++) {
            int light = (i + vt) % (c + 1);
            int amount = count[i][vx2][vy2] - count[i][vx1 - 1][vy2] - count[i][vx2][vy1 - 1] + count[i][vx1 - 1][vy1 - 1];
            result += amount * light;
        }
        return result;
    }
}