Problem: Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.

Input

The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).

Output

On the single line print the result with an absolute error of no more than 10 - 9.

Examples

Input

1 10 1 10 2


Output

0.320000000000


Input

5 6 8 10 1


Output

1.000000000000

Note

Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>

In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.

In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number.

Solution (in JAVA):

Solution:


import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class LuckyProb {


	static ArrayList<Integer> lucky = new ArrayList();
	public static void gen(long x)
	{
		if(x >= (int)(10e9))
			return;
		if(x != 0)
			lucky.add((int) x);
		gen(x*10L + 4);
		gen(x*10L + 7);
	}

	public static void main(String[]args)throws Throwable
	{
		//
		Scanner sc = new Scanner(System.in);
		gen(0);
		Collections.sort(lucky);
		int L1 = sc.nextInt(), R1 = sc.nextInt(), L2 = sc.nextInt(), R2 = sc.nextInt(), k = sc.nextInt();
		long total = 0;
		for(int i = 0 ; i < 1022 ; ++i)
		{
			if(i + k - 1 >= 1022)
				break;
			int min = lucky.get(i);
			int max = lucky.get(i + k - 1);
			
			int x = 0;
			if(L1 <= min && R2 >= max)
			{
		
				++x;
				// it can be covered
				long before = Math.min(min,R1) - Math.max(L1,(i == 0 ? L1 : lucky.get(i - 1) + 1)) + 1;
				long after = Math.min(R2, i + k == 1022 ? R2 : lucky.get(i + k) - 1) - Math.max(max,L2) + 1;
				if(before < 0 || after < 0)
					continue;
				total += (long)(1L*after * before);
				//System.out.println(before +" "+after);
				
		
			}
			if(L2 <= min && R1 >= max)
			{
				// it can be covered
				++x;
				//System.out.println("in");
				long before = Math.min(min,R2) - Math.max(L2,(i == 0 ? L2 : lucky.get(i - 1) + 1)) + 1;
				long after = Math.min(R1, i + k == 1022 ? R1 : lucky.get(i + k) - 1) -  Math.max(max,L1) + 1;
				if(before < 0 || after < 0)
					continue;
				total += (long)(1L*after * before);
				
			}
			if(x == 2 && k == 1)
				--total;
			

		}
		
		long all = (long)((long)R1 - L1 + 1) * ((long)R2 - L2 + 1);
		double ans = (total*1.0)/(all*1.0);
		System.out.printf("%.12f",Math.min(1,ans));

	}
}