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 {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);

        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int q = Integer.parseInt(st.nextToken());
        int c = Integer.parseInt(st.nextToken());

        int[][][] stars = new int[c+1][101][101];

        for(int i = 0; i<n; i++){
            st = new StringTokenizer(br.readLine());
            int x = Integer.parseInt(st.nextToken());
            int y = Integer.parseInt(st.nextToken());
            int s = Integer.parseInt(st.nextToken());
            stars[s][y][x]++;
        }

        for(int i = 0; i<=c; i++){
            for(int j = 1; j<101; j++){
                for(int k = 1; k<101; k++){
                    stars[i][j][k] += stars[i][j-1][k] + stars[i][j][k-1] - stars[i][j-1][k-1];
                }
            }
        }

        for(int i = 0; i<q; i++){
            st = new StringTokenizer(br.readLine());
            int t = Integer.parseInt(st.nextToken());
            int x = Integer.parseInt(st.nextToken());
            int y = Integer.parseInt(st.nextToken());
            int x2 = Integer.parseInt(st.nextToken());
            int y2 = Integer.parseInt(st.nextToken());

            long brightnessCount = 0;

            for(int j = 0; j<=c; j++){
                long starCount = 0;
                starCount = stars[j][y2][x2] + stars[j][y-1][x-1] - stars[j][y2][x-1] - stars[j][y-1][x2];
                brightnessCount += starCount * ( (j + t) % (c+1));
            }

            pw.println(brightnessCount);
        }

        pw.close();
    }

}