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.InputStreamReader;

/**
 * Created by shikhar on 31/07/17.
 */
public class Main {
    public static void main(String[] args)throws IOException
    {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String[] arr = in.readLine().trim().split(" ");
        int N = Integer.parseInt(arr[0]);
        int Q = Integer.parseInt(arr[1]);
        int C = Integer.parseInt(arr[2]);
        int dp[][][] = new int[11][102][102];
        for(int i =0; i<N; i++) {
            arr = in.readLine().trim().split(" ");
            dp[Integer.parseInt(arr[2])][Integer.parseInt(arr[0])+1][Integer.parseInt(arr[1])+1]++;
        }

        for(int i = 0; i<=C; i++)
            for(int j = 1 ; j<=101; j++)
                for(int k = 1; k<=101; k++)
                    dp[i][j][k]+=dp[i][j-1][k]+dp[i][j][k-1]-dp[i][j-1][k-1];

        for(int i =0; i<Q; i++)
        {
            arr = in.readLine().trim().split(" ");
            int time = Integer.parseInt(arr[0]);
            int x1 = Integer.parseInt(arr[1]);
            int y1 = Integer.parseInt(arr[2]);
            int x2 = Integer.parseInt(arr[3]);
            int y2 = Integer.parseInt(arr[4]);

            long ans = 0l;
            for(int c = 0; c<=C; c++) {
                ans += (dp[c][x2 + 1][y2 + 1] - dp[c][x1][y2 + 1] - dp[c][x2 + 1][y1] + dp[c][x1][y1]) * ((c + time) % (C + 1));
            }
        System.out.println(ans);
        }
    }
}
