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.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class My_c
{
	//7araaaaaaaaaaaaaaaaaaaam elbug di
	public static void main(String[] args) throws IOException
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int q = sc.nextInt();
		int c = sc.nextInt();
		long[][] arr = new long[101][101];
		int[][] cnt = new int[101][101];
		long[][] cum = new long[101][101];
		int[][] cumCnt = new int[101][101];
		int[][][] cntMoreThan = new int[c+1][101][101];
		for(int i=0;i<n;i++)
		{
			int x = sc.nextInt();
			int y = sc.nextInt();
			int s = sc.nextInt();
			arr[x][y]+=s;
			cnt[x][y]++;

			for(int k=0;k<c+1;k++)
				if(s>k)
					cntMoreThan[k][x][y]++;
		}
		for(int i=1;i<=100;i++)
		{
			for(int j=1;j<=100;j++)
			{
				cum[i][j] = arr[i][j];
				cum[i][j]+=cum[i-1][j];
				cum[i][j]+=cum[i][j-1];
				cum[i][j]-=cum[i-1][j-1];
				cumCnt[i][j] = cnt[i][j];
				cumCnt[i][j]+=cumCnt[i-1][j];
				cumCnt[i][j]+=cumCnt[i][j-1];
				cumCnt[i][j]-=cumCnt[i-1][j-1];
			}
		}
		int[][][] cumMoreThan = new int[c+1][101][101];
		for (int k = 0; k < cntMoreThan.length; k++)
		{
			for(int i=1;i<=100;i++)
			{
				for(int j=1;j<=100;j++)
				{
					cumMoreThan[k][i][j] = cntMoreThan[k][i][j];
					cumMoreThan[k][i][j]+=cumMoreThan[k][i-1][j];
					cumMoreThan[k][i][j]+=cumMoreThan[k][i][j-1];
					cumMoreThan[k][i][j]-=cumMoreThan[k][i-1][j-1];
				}
			}
		}
		PrintWriter pw = new PrintWriter(System.out);
		while(q-->0)
		{
			int t = sc.nextInt();
			t = t%(c+1);
			int x1 =sc.nextInt();
			int y1 =sc.nextInt();
			int x2 =sc.nextInt();
			int y2 =sc.nextInt();
			long sum = cum[x2][y2]+cum[x1-1][y1-1] -cum[x2][y1-1] - cum[x1-1][y2];
			int cntt = cumCnt[x2][y2]+cumCnt[x1-1][y1-1] -cumCnt[x1-1][y2]-cumCnt[x2][y1-1];
			int threshold = c-t;
			int cntMore = cumMoreThan[threshold][x2][y2]+cumMoreThan[threshold][x1-1][y1-1]
					-cumMoreThan[threshold][x1-1][y2] -cumMoreThan[threshold][x2][y1-1];
			long ans = sum + 1l*cntt * t - 1l*cntMore * (c+1);
			pw.println(ans);
		}
		pw.flush();
	}

	static class Scanner{
		StringTokenizer st;BufferedReader br;
		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}
		public Scanner(String s) throws FileNotFoundException{	br = new BufferedReader(new FileReader(new File(s)));}
		public String next() throws IOException 
		{while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());return st.nextToken();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		public long nextLong() throws IOException {return Long.parseLong(next());}
		public String nextLine() throws IOException {return br.readLine();}
		public double nextDouble() throws IOException {return Double.parseDouble(next());}
		public boolean ready() throws IOException {return br.ready();}
	}
}