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.*;
import java.math.*;

public class C{
	
	static int n,q,c,a[][][];
	static long s[][][];
	static void MainMethod()throws Exception{
		n=reader.nextInt();
		q=reader.nextInt();
		c=reader.nextInt();
		c++;
		a=new int[c][102][102];s=new long[c][102][102];
		int t,i,j,k,x,y,x1,x2,y1,y2;
		for (i=0;i<c;i++)for(j=0;j<102;j++){
			Arrays.fill(a[i][j],0);
			Arrays.fill(s[i][j],0);
		}
		for (i=0;i<n;i++){ 
			x=reader.nextInt();
			y=reader.nextInt();
			k=reader.nextInt();
			a[0][x][y]+=k;
			for (j=1;j<c;j++)
				a[j][x][y]+=(k+j)%c;
		}
		for (k=0;k<c;k++)
			for (x=1;x<102;x++)for (y=1;y<102;y++){
				s[k][x][y]=s[k][x-1][y]+s[k][x][y-1]-s[k][x-1][y-1]+a[k][x][y];
			}
		long res;
		for (i=0;i<q;i++){
			t=reader.nextInt()%c;
			x1=reader.nextInt();
			y1=reader.nextInt();
			x2=reader.nextInt();
			y2=reader.nextInt();
			res=s[t][x2][y2]-s[t][x1-1][y2]-s[t][x2][y1-1]+s[t][x1-1][y1-1];
			printer.println(res);
		}
		
	}
	public static void main(String[] args)throws Exception{
		MainMethod();
		printer.close();
	}
	static void halt(){
		printer.close();
		System.exit(0);
	}
	static PrintWriter printer=new PrintWriter(new OutputStreamWriter(System.out));
	static class reader{
		static BufferedReader bReader=new BufferedReader(new InputStreamReader(System.in));
		static StringTokenizer token=new StringTokenizer("");
		static String readNextLine() throws Exception{
			return bReader.readLine();
		}
		static String next() throws Exception{
			while (token.hasMoreTokens()==false){
				token=new StringTokenizer(bReader.readLine());
			}
			return token.nextToken();
		}
		static int nextInt()throws Exception{
			while (token.hasMoreTokens()==false){
				token=new StringTokenizer(bReader.readLine());
			}
			return Integer.parseInt(token.nextToken());
		}
		static long nextLong()throws Exception{
			while (token.hasMoreTokens()==false){
				token=new StringTokenizer(bReader.readLine());
			}
			return Long.parseLong(token.nextToken());
		}
		static double nextDouble()throws Exception{
			while (token.hasMoreTokens()==false){
				token=new StringTokenizer(bReader.readLine());
			}
			return Double.parseDouble(token.nextToken());
		}
	}
	static class MyMathCompute{
		static long [][] MatrixMultiplyMatrix(long [][] A, long [][] B, long mod) throws Exception{
			int n=A.length, m=B[0].length; 
			int p=A[0].length;
			int i,j,k;
			if (B.length!=p) throw new Exception("invalid matrix input");
			long [][] res=new long [n][m];
			for (i=0;i<n;i++) for (j=0;j<m;j++){
				if (A[i].length!=p) throw new Exception("invalid matrix input");
				res[i][j]=0;
				for (k=0;k<p;k++)
					res[i][j]=(res[i][j]+((A[i][k]*B[k][j])% mod))% mod;
			}
			return res;
		}
		static double [][] MatrixMultiplyMatrix(double [][] A, double [][] B ) throws Exception{
			int n=A.length, m=B[0].length; 
			int p=A[0].length;
			int i,j,k;
			if (B.length!=p) throw new Exception("invalid matrix input");
			double [][] res=new double [n][m];
			for (i=0;i<n;i++) for (j=0;j<m;j++){
				if (A[i].length!=p) throw new Exception("invalid matrix input");
				res[i][j]=0;
				for (k=0;k<p;k++)
					res[i][j]=res[i][j]+(A[i][k]*B[k][j]);
			}
			return res;
		}
		static long [][] MatrixPow(long [][] A,long n, long mod) throws Exception{
			if (n==1) return A;
			long [][] res=MatrixPow(A, n/2, mod);
			res=MatrixMultiplyMatrix(res, res, mod);
			if ((n % 2) == 1) res=MatrixMultiplyMatrix(A,res, mod);		
			return res;
		}
		static double [][] MatrixPow(double [][] A,long n) throws Exception{
			if (n==1) return A;
			double[][] res=MatrixPow(A, n/2);
			res=MatrixMultiplyMatrix(res, res);
			if ((n % 2) == 1) res=MatrixMultiplyMatrix(A,res);		
			return res;
		}
		static long pow(long a,long n,long mod){
			a= a % mod;
			if (n==0) return 1;
			long k=pow(a,n/2,mod);
			if ((n % 2)==0) return ((k*k)%mod);
			else return (((k*k) % mod)*a) % mod;
		}
		static double pow(double a,long n){
			if (n==0) return 1;
			double k=pow(a,n/2);
			if ((n % 2)==0) return (k*k);
			else return (((k*k) )*a) ;
		}
	}
}
