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:
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF109B extends PrintWriter {
	CF109B() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF109B o = new CF109B(); o.main(); o.flush();
	}

	static final int A = 1000000000;
	long lucky(int i) {
		return i == 1 ? 0 : lucky(i / 2) * 10 + (i % 2 == 0 ? 4 : 7);
	}
	int intersect(int al, int ar, int bl, int br) {
		int l = Math.max(al, bl);
		int r = Math.min(ar, br);
		return l <= r ? r - l + 1 : 0;
	}
	void main() {
		int ul = sc.nextInt();
		int ur = sc.nextInt();
		int vl = sc.nextInt();
		int vr = sc.nextInt();
		int k = sc.nextInt();
		int[] qq = new int[1024];
		int cnt = 0;
		qq[cnt++] = 0;
		long a;
		for (int i = 2; (a = lucky(i)) <= A; i++)
			qq[cnt++] = (int) a;
		qq[cnt++] = A + 1;
		long cnt_ = 0;
		for (int i = 1, j; (j = i + k - 1) < cnt - 1; i++) {
			int pl = qq[i - 1] + 1;
			int pr = qq[i];
			int ql = qq[j];
			int qr = qq[j + 1] - 1;
			cnt_ += (long) intersect(ul, ur, pl, pr) * intersect(vl, vr, ql, qr);
			cnt_ += (long) intersect(ul, ur, ql, qr) * intersect(vl, vr, pl, pr);
			if (i == j && ul <= qq[i] && qq[i] <= ur && vl <= qq[i] && qq[i] <= vr)
				cnt_--;
		}
		printf("%.9f\n", (double) cnt_ / (ur - ul + 1) / (vr - vl + 1));
	}
}