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 Ishu
{
    static Scanner scan = new Scanner(System.in);
    static void tc()
    {
    int n = scan.nextInt();
    int q = scan.nextInt();
    int c = scan.nextInt();
    int[][][] sum = new int[101][101][11];
    int[][][] dp = new int[101][101][11];

    int i,j;
    for(i=0;i<n;++i)
        {
        int x = scan.nextInt();
        int y = scan.nextInt();
        int s = scan.nextInt();
        sum[y][x][s]++;
        }
    
    for(i=1;i<101;++i)
        for(j=1;j<101;++j)
            {
            int[] nn = new int[11];
                    
            for(int k=0;k<j;++k)
                {
                for(int ctr = 0;ctr <= 10;++ctr)
                    nn[ctr] += sum[i][k][ctr];
                }
                    
            for(int k=0;k<i;++k)    
                {   
                for(int ctr = 0;ctr <= 10;++ctr)
                    nn[ctr] += sum[k][j][ctr];
                }
            for(int k=0;k<=10;++k)    
                dp[i][j][k] = sum[i][j][k] + dp[i-1][j-1][k] + nn[k];
            }
            
    for(i=0;i<q;++i)
        { 
        long t = scan.nextLong();
        int y1 = scan.nextInt();
        int x1 = scan.nextInt();
        int y2 = scan.nextInt();
        int x2 = scan.nextInt();
        long ans = 0l;
       
        
       
        int[] nn = new int[11];
        
        for(int k=0;k<=10;++k)
            {
            nn[k] = dp[x2][y2][k] - dp[x1-1][y2][k] - dp[x2][y1-1][k] + dp[x1-1][y1-1][k];
            ans += nn[k] * ((k + t) % (c + 1));
            }
        
        System.out.println(ans);
        }
    }
    public static void main(String[] args)
    {
    int t = 1;
    //t = scan.nextInt();
    while(t-- > 0)
        tc();
    }
}
