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.BigDecimal;
import java.math.BigInteger;
import java.text.DecimalFormat;
import java.text.NumberFormat;
import java.math.*;
import static java.lang.Math.*;
import java.util.*;

import javax.swing.plaf.synth.SynthSeparatorUI;

public class Main {
    public  static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0)
        {
            count += n & 1;
            n >>= 1;
        }
        return count;
    }
    public static int n,a,b,c;
    public static int dp[]=new int[4001];
    public static int calc(int i){
        if(i==0)return 0;
        if(dp[i]!=-1)return dp[i];
        int v=Integer.MIN_VALUE;
        if(i >= a) v = max(v, calc(i-a)+1);
        if(i >= b) v = max(v, calc(i-b)+1);
        if(i >= c) v = max(v, calc(i-c)+1);
        return dp[n] = v;
    }

    public static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

    public static PrintWriter pw=new PrintWriter(System.out);
    public static void main(String[] args) throws IOException {
        // TODO Auto-generated method stub
        Scanner S = new  Scanner(System.in);
        int n=S.nextInt(); int q=S.nextInt();  int c=S.nextInt();
        int a[][][]=new int [101][101][c+1];
        while(n-->0) ++a[S.nextInt()][S.nextInt()][S.nextInt()];
        int x=0, y=0, b=0;
        for(x = 1; x < 101; x++)
            for(y = 1; y < 101; y++)
                for(b = 0; b <= c; b++)
                    a[x][y][b] += a[x - 1][y][b] + a[x][y - 1][b] - a[x - 1][y - 1][b];
        long t, sum, cur;
        int x1, y1, x2, y2;
        while(q-- > 0) {
            t = S.nextLong();
            x1 = S.nextInt();
            y1 = S.nextInt();
            x2 = S.nextInt();
            y2 = S.nextInt();
            sum = 0;
            for(b = 0; b <= c; b++) {
                if(t + b <= c)
                    cur = t + b;
                else {
                    cur = (t % (c + 1)) + b;
                    if(cur > c)
                        cur -= c + 1;
                }
                sum += (a[x2][y2][b] - a[x1 - 1][y2][b] - a[x2][y1 - 1][b] + a[x1 - 1][y1 - 1][b]) * cur;
            } 	
        
            System.out.println(sum);
        }
    }

}