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.Scanner;

public class B109 {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long pl = sc.nextLong();
        long pr = sc.nextLong();
        long vl = sc.nextLong();
        long vr = sc.nextLong();
        int k = sc.nextInt();
        long n = (vr-vl+1) * (pr-pl+1);
        long m = 0;
        ArrayList<Integer>luckly = new ArrayList<Integer>();
        luckly.add(0);
        for (int i = 1; i <= 9; i++) {
            for (int j = 0; j < (1 << i); j++) {
                int s = 0;
                for (int j2 = i-1; j2 >= 0; j2--) {
                    s *= 10;
                    if ((j & (1 << j2))==0)
                        s += 4;
                    else
                        s += 7;
                }
                luckly.add(s);
            }
        }
        luckly.add((int) 1e9);
        for (int i = 1; i < luckly.size()-k; i++) {
            long k1 = 0,k2 = 0,k3 = 0,k4 = 0;
            if (vl <= luckly.get(i) && vr > luckly.get(i-1)) {
                k1 = Math.min(vr, luckly.get(i))-Math.max(luckly.get(i-1)+1, vl)+1;
            }
            if (vr >= luckly.get(i+k-1) && vl < luckly.get(i+k)) {
                k2 = Math.min(vr, luckly.get(i+k)-1)-Math.max(vl, luckly.get(i+k-1))+1;;
            }
            if (pl <= luckly.get(i) && pr > luckly.get(i-1)) {
                k3 = Math.min(pr, luckly.get(i))-Math.max(luckly.get(i-1)+1, pl)+1;
            }
            if (pr >= luckly.get(i+k-1) && pl < luckly.get(i+k)) {
                k4 = Math.min(pr, luckly.get(i+k)-1)-Math.max(pl, luckly.get(i+k-1))+1;;
            }
            m += k1*k4+k2*k3;
            if (k==1 && k1 != 0 && k2 != 0 && k3 != 0 && k4 != 0) {
                m--;
            }
        }
        System.out.println(Math.min(1,m*1.0/n));
    }
}