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.Arrays;
import java.util.Scanner;


public class B109 {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long a = sc.nextInt();
        long b = sc.nextInt();
        long c = sc.nextInt();
        long d = sc.nextInt();
        int k = sc.nextInt();
        int m = 0;
        long[] mas = new long[1024];
        for (int i = 1; i <= 9; i++) {
            for (int j = 0; j < (1 << i); j++) {
                String s = Integer.toBinaryString(j);
                while (s.length() < i)
                    s = "0"+s;
                String ss = "";
                for (int l = 0; l < i; l++) {
                    if (s.charAt(l) == '1')
                        ss += '7';
                    else
                        ss += '4';
                }
                mas[++m] = Integer.parseInt(ss);
            }
        }
        long ans = 0;
        mas[++m] = (long) 1e9;
        for (int i = 1; i < m; i++) {
            if (i+k > m)
                break;
            long x1 = (long) Math.max(mas[i-1]+1, a);
            long y1 = (long) Math.min(mas[i], b);
            long x2 = (long) Math.max(mas[i+k-1], c);
            long y2 = (long) Math.min(mas[i+k]-1, d);
//          System.out.println(i+" -- "+x1+" "+y1+" - "+x2+" "+y2);
            if (x1 > y1 || x2 > y2)
                continue;
            else {
                ans += (y1 - x1 + 1) * (y2 - x2 + 1);
            }
        }
        for (int i = 1; i < m; i++) {
            if (i+k > m)
                break;
            long x1 = (long) Math.max(mas[i-1]+1, c);
            long y1 = (long) Math.min(mas[i], d);
            long x2 = (long) Math.max(mas[i+k-1], a);
            long y2 = (long) Math.min(mas[i+k]-1, b);
            if (x1 > y1 || x2 > y2)
                continue;
            else {
                ans += (y1 - x1 + 1) * (y2 - x2 + 1);
                if (k == 1 && y1 == x2 && Arrays.binarySearch(mas, y1) > 0)
                    ans--;
            }
        }
        double p = ans*1.0/((b-a+1)*(d-c+1));
        System.out.printf("%.9f",p);
    }

}