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.math.BigInteger;
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.util.*;
import java.util.Map.Entry;
import java.util.regex.Pattern;

import javax.management.relation.RelationServiceNotRegisteredException;


public class hehehe {

	static int count=0;
	static int flag=0;
	static double eps=(double)1e-6;
	static long mod=(long)1e9+7;
	static int maxcount=0;
	static int index=0;
	static int fac[]=new int[10];
	static int end_time[]=new int[100001];
	static int bit[]=new int[100001];
	static int vis_time[]=new int[100001];
	static int flat_tree[]=new int[200001];
	static ArrayList<ArrayList<Integer>> tree=new ArrayList<>();
	static ArrayList<ArrayList<Integer>> table=new ArrayList<>();
	static int traverser[]=new int[100001];
	static int par[]=new int[100001];
	static boolean vis[]=new boolean[100001];
	static int tim = 0;

	//li, ri and index are stored in queries vector
	//in that order, as the sort function will use
	//the value li for comparison
	static ArrayList< pair1 > queries=new ArrayList<>();

	//ans[i] stores answer to ith query
	static int ans[]=new int[100001];
	static HashSet<String> hm;

	public static void main(String args[]){
		InputReader in = new InputReader(System.in);
		OutputStream outputStream = System.out;
		PrintWriter out = new PrintWriter(outputStream);
		Scanner sc=new Scanner(System.in);
		//----------My Code----------
		int n=in.nextInt();
		int q=in.nextInt();
		int c=in.nextInt();
		
		int a[][][]=new int[11][101][101];
		for(int i=0;i<n;i++){
			int xi=in.nextInt();
			int yi=in.nextInt();
			int si=in.nextInt();
			
			a[si][xi][yi]+=1;
			
		}
		//int a[][][]=new int[11][101][101];
		for(int i=0;i<=10;i++){
			
			for(int j=1;j<101;j++){
				
				for(int k=1;k<101;k++){
					
					a[i][j][k]+=a[i][j][k-1]+a[i][j-1][k]-a[i][j-1][k-1];
					
				}
			}
			
		}
		for(int i=0;i<q;i++){
			int t1=in.nextInt();
			int x1=in.nextInt();
			int y1=in.nextInt();
			int x2=in.nextInt();
			int y2=in.nextInt();
			//System.out.println(x1+""+y1+""+x2+""+y2);
			long ans=0;
			for(int k=0;k<=(c);k++){
				//System.out.println(a[k][x2][y2]+" "+a[k][x2][y1]+""+a[k][x1][y2]+""+a[k][x1][y1]);
				ans+=((k+t1)%(c+1))*(a[k][x2][y2]-a[k][x2][y1-1]-a[k][x1-1][y2]+a[k][x1-1][y1-1]);
				
			}
			
			
			System.out.println(ans);
		
		
		}
		
		
		
		
		
		//-- -------------The End----------------- 	-
		out.close();
	}
	
	static int rec(String ans,int length){
		if(length<=2){
			return 0;
		}
		else{
			
			int flag=0;
		char start=ans.charAt(0);
		for(int i=1;i<ans.length()-1;i++){
       	 if(start=='0'&&ans.charAt(i+1)=='0'){
                if(ans.charAt(i)=='0'){
                    flag=1;
                  //  System.out.println(ans.substring(0, i)+ans.substring(i+1));
                   count=count^rec(ans.substring(0, i)+ans.substring(i+1),ans.length()-1);
                }
                else if (ans.charAt(i)=='1'){
                     flag=1;
                   //  System.out.println(ans.substring(0, i)+ans.substring(i+1));
                	 count=count^rec(ans.substring(0, i)+ans.substring(i+1),ans.length()-1);
                 
                }
               
            }
       	 start=ans.charAt(i);
        }
		if(flag==0)
			return count^1;
		return count;
		}	
	}
	
	static int recursiveBinarySearch(int[] sortedArray, int start, int end, int key) {

		while (start < end) {
			int mid = start + (end - start) / 2; 
			if (key < sortedArray[mid]) {
				end=mid-1;

			} else if (key >=sortedArray[mid]) {
				start=mid+1;
			}
		}
		return (start); 
	}

	static long modulo(long a,long b,long c) {
		long x=1;
		long y=a;
		while(b > 0){
			if(b%2 == 1){
				x=(x*y)%c;
			}

			y = (y*y)%c; // squaring the base
			b /= 2;
		}
		return  x%c;
	}


	static long gcd(long x, long y)

	{
		if(x==0||y==0)
			return 0;
		long r=0, a, b;
		a = (x > y) ? x : y; // a is greater number
		b = (x < y) ? x : y; // b is smaller number
		r = b;
		while(a % b != 0)
		{
			r = a % b;
			a = b;
			b = r;
		}
		return r;
	}
	static class Pair implements Comparable<Pair> {
		int x;
		int y;
		//int i;
		Pair(int  xx,int yy){

			x=xx;
			y=yy;
			//i=ii;
		}
		@Override
		public int compareTo(Pair arg0) {

			return Integer.compare(10*x+y,10*y+x);


		}

	}
	static class pair1 implements Comparable<pair1>{
		Pair p;
		int x;
		pair1(Pair p,int x){
			this.p=p;
			this.x=x;
		}
		public int compareTo(pair1 arg0) {



			return Integer.compare(p.x, arg0.p.x);


		}

	}
	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream inputstream) {
			reader = new BufferedReader(new InputStreamReader(inputstream));
			tokenizer = null;
		}

		public String nextLine(){
			String fullLine=null;
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					fullLine=reader.readLine();
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
				return fullLine;
			}
			return fullLine;
		}
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
} 