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:
/*
 * ACM Code => cf110d.java
 * Written by Sam X
 * Date: 三月, 08, 2019
 * Time: 14:27
 */
import java.util.*;
import java.math.*;

public class cf110d
{
    static ArrayList<Long> vec = new ArrayList<>();
    static void dfs(long x)
    {
        if(x>1e9) return;
        if(x*10+4<1e9)
        {
            vec.add(x*10+4);
            dfs(x*10+4);
        }
        if(x*10+7<1e9)
        {
            vec.add(x*10+7);
            dfs(x*10+7);
        }
    }
    static final long contain(long x1, long y1, long x2, long y2)
    {
        return Math.max(Math.min(y1,y2)-Math.max(x1,x2)+1,0l);
    }
    public static void main(String args[])
    {
        Scanner cin = new Scanner(System.in);
        long pl = cin.nextLong(), 
             pr = cin.nextLong(), 
             vl = cin.nextLong(),
             vr = cin.nextLong(),
             k = cin.nextLong();
    
        dfs(0);
        vec.add(0l);
        vec.add((long)1e9);

        Collections.sort(vec);
        /*
        for(long x: vec)
        {
            System.out.print(x+" ");
        }
        System.out.println();
        */
        long ans=0;

        int sz=vec.size()-2;
        for(int i=1; i<=sz-k+1; ++i)
        {
            int j=i+(int)k-1;
            ans+=contain(vec.get(i-1)+1, vec.get(i), pl, pr)*contain(vec.get(j), vec.get(j+1)-1, vl, vr);
            if(vec.get(i)>pr) break;
        }

        for(int i=1; i<=sz-k+1; ++i)
        {
            int j=i+(int)k-1;
            ans+=contain(vec.get(i-1)+1, vec.get(i), vl, vr)*contain(vec.get(j), vec.get(j+1)-1, pl, pr);
            if(vec.get(i)>vr) break;
        }

        if(k==1)
        {
            for(int i=1; i<=sz; ++i)
            {
                if(contain(vec.get(i), vec.get(i), pl, pr)!=0 &&
                   contain(vec.get(i), vec.get(i), vl, vr)!=0) ans--;
            }
        }

        System.out.printf("%.12f\n", (double)ans/(vr-vl+1)/(pr-pl+1));
        cin.close();
    }
}