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.*;
import java.math.*;
import java.util.concurrent.*;

public final class star_sky
{
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	static FastScanner sc=new FastScanner(br);
    static PrintWriter out=new PrintWriter(System.out);
	static Random rnd=new Random();
	static Node[] arr;
	static int maxn=(int)(105);
	
    public static void main(String args[]) throws Exception
    {
		int n=sc.nextInt(),q=sc.nextInt(),c=sc.nextInt();arr=new Node[11];
		
		for(int i=0;i<11;i++)
		{
			arr[i]=new Node();
		}
		
		for(int i=0;i<n;i++)
		{
			int y=sc.nextInt(),x=101-sc.nextInt(),s=sc.nextInt();
			
			arr[s].cnt[x][y]++;
		}
		
		for(int i=0;i<11;i++)
		{
			for(int j=0;j<maxn;j++)
			{
				long curr=0;
				
				for(int k=0;k<maxn;k++)
				{
					curr+=arr[i].cnt[j][k];
					
					long curr2=(j>0?arr[i].a[j-1][k]:0);
					
					arr[i].a[j][k]=curr+curr2;
				}
			}
		}
		
		while(q>0)
		{
			
			int t=sc.nextInt()%(c+1),y1=sc.nextInt(),x2=101-sc.nextInt(),y2=sc.nextInt(),x1=101-sc.nextInt();
			
			long res=0;
			
			for(int i=0;i<11;i++)
			{
				long curr_val=(t+i)%(c+1);
				
				long val1=arr[i].a[x2][y2],val2=arr[i].a[x2][y1-1],val3=arr[i].a[x1-1][y2],val4=arr[i].a[x1-1][y1-1];
				
				long tot=val1-val2-val3+val4;
				
				res=(res+(tot*curr_val));
			}
			
			out.println(res);
			
			q--;
		}
		
		out.close();
    }
}

class Node
{
	static int maxn=(int)(105);
	
	long[][] a=new long[maxn][maxn],cnt=new long[maxn][maxn];
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}