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.util.* ;
import java.io.BufferedReader ;
import java.io.InputStreamReader ;

public class StarSky
{
    static boolean debug = true ;
    public static void main(String args[]) throws Exception
    {
        BufferedReader bro = new BufferedReader(new InputStreamReader(System.in)) ;
        String[] S = bro.readLine().split(" ") ;
        
        int[] hasPoint = new int[(int)1e6+10000+11] ;
        int n = Integer.parseInt(S[0]) ;
        int q = Integer.parseInt(S[1]) ;
        int c = Integer.parseInt(S[2]) ;
        for(int i=0;i<n;i++)
        {
            S = bro.readLine().split(" ") ;
            int x = Integer.parseInt(S[0]) ;
            int y = Integer.parseInt(S[1]) ;
            int s = Integer.parseInt(S[2]) ;
            // H.add((int)1e4*x+100*y+x) ;
            hasPoint[(int)1e4*x+100*y+s]++ ;
        }
        int[][][] dp = new int[101][101][c+1] ;
        for(int p=0;p<=c;p++)
        {
            for(int x=1;x<=100;x++)
            {
                for(int y=1;y<=100;y++)
                {
                    dp[x][y][p] = dp[x-1][y][p]+dp[x][y-1][p]-dp[x-1][y-1][p]+ hasPoint[(int)1e4*x+100*y+p] ;
                 
                }
            }
        }
        for(int i=0;i<q;i++)
        {
            S = bro.readLine().split(" ") ;
            int t = Integer.parseInt(S[0]) ;
            int x1 = Integer.parseInt(S[1]) ;
            int y1 = Integer.parseInt(S[2]) ;
            int x2 = Integer.parseInt(S[3]) ;
            int y2 = Integer.parseInt(S[4]) ;
            // System.out.println((dp[x2][y2][p]-dp[x1-1][y2][p]-dp[x2][y1-1][p]+dp[x1-1][x2-1][p])) ;
            int sum = 0 ;
            
            for(int p=0;p<=c;p++)
            {
                int val = (t+p)%(c+1) ;
                int points = dp[x2][y2][p]-dp[x1-1][y2][p]-dp[x2][y1-1][p]+dp[x1-1][y1-1][p] ;
                sum += val*points ;
            }
            System.out.println(sum);
        }

    }
}