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.*;

public class cf109b {
	static ArrayList<Long> lucky = new ArrayList<Long>();
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		genLucky(0,11);
		Collections.sort(lucky);
		long pl = in.nextLong();
		long pr = in.nextLong();
		long vl = in.nextLong();
		long vr = in.nextLong();
		int k = in.nextInt();
		double ret = 0;
		for(int i=0; i<lucky.size()-k-2; i++) {
			//assume p is lower
			//probability that p got in the interval (lucky[i],lucky[i+1]]
			long lo = Math.max(lucky.get(i)+1, pl);
			long hi = Math.min(lucky.get(i+1), pr);
			double p = Math.max(0,(hi-lo+1.)/(pr-pl+1));
			//probability that v got in the interval [lucky[i+k],lucky[i+k+1)
			lo = Math.max(lucky.get(i+k),vl);
			hi = Math.min(lucky.get(i+k+1)-1,vr);
			double v = Math.max(0,(hi-lo+1.)/(vr-vl+1));
			ret += p*v;
			
			//assume v is lower
			//probability that v got in the interval (lucky[i],lucky[i+1]]
			lo = Math.max(lucky.get(i)+1, vl);
			hi = Math.min(lucky.get(i+1), vr);
			v = Math.max(0,(hi-lo+1.)/(vr-vl+1));
			//probability that p got in the interval [lucky[i+k],lucky[i+k+1)
			lo = Math.max(lucky.get(i+k),pl);
			hi = Math.min(lucky.get(i+k+1)-1,pr);
			p = Math.max(0,(hi-lo+1.)/(pr-pl+1));
			ret += p*v;
			
			//remove double counting (probability they got the same)
			if(k==1) {
				lo = Math.max(lucky.get(i), vl);
				hi = Math.min(lucky.get(i), vr);
				v = Math.max(0,(hi-lo+1.)/(vr-vl+1));
				lo = Math.max(lucky.get(i),pl);
				hi = Math.min(lucky.get(i),pr);
				p = Math.max(0,(hi-lo+1.)/(pr-pl+1));
				ret -= p*v;
			}
		}
		System.out.printf("%.12f%n", ret);
	}
	static void genLucky(long cur, long left) {
		if(left == 0) return;
		lucky.add(cur);
		genLucky(cur*10+4,left-1);
		genLucky(cur*10+7,left-1);
	}
}