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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class D 
{
    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    public static StringTokenizer inp;
    public static void main(String[] args)  throws IOException
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        
        int ln[] = new int[1026];
        ln[0] = 0 ; ln[1023] = 2000000000;
        int ind = 1;
        for(int i = 2 ; i < 1024 ; i++)
        {
            char s[] = Integer.toBinaryString(i).toCharArray();
            for(int j = 1 ; j < s.length ; j++) s[j] = (s[j]=='0'?'4':'7');
            ln[ind++] = Integer.parseInt(String.valueOf(s).substring(1));
        }
        inp = new StringTokenizer(br.readLine());
        int pl = nextInt(),pr = nextInt() , vl=nextInt() , vr = nextInt() , k = nextInt();
        long ans = 0;
        int l = 1, r = k;
        for(; r <= 1022 ; r++,l++)
        {
            if(vl>ln[l])continue;
            if(ln[r] > pr || ln[l-1] > vr)break;
            ans+=(long)Math.max(0, (Math.min(ln[l], vr)- Math.max(vl, ln[l-1]+1) + 1))*Math.max(0, (Math.min(pr, ln[r+1]-1)- Math.max(ln[r], pl) + 1));
        }
        l=1;r=k;
        for(; r <= 1022 ; r++,l++)
        {
            if(pl>ln[l])continue;
            if(ln[r] > vr || ln[l-1] > pr)break;
            ans+=(long)Math.max(0, (Math.min(ln[l], pr)- Math.max(pl, ln[l-1]+1) + 1))*Math.max(0, (Math.min(vr, ln[r+1]-1)- Math.max(ln[r], vl) + 1));
        }
        if(k==1)
        {
            for(int i = 1; i <= 1022 ; i++)
                if(pl<=ln[i] && ln[i]<=pr && vl<=ln[i]&&ln[i]<=vr)ans--;
        }
        double res = (double)(ans)/((long)(pr-pl+1)*(vr-vl+1));
        System.out.printf("%.12f\n",Math.min(res, 1.0));
    }
    
    public static int nextInt(){return Integer.parseInt(inp.nextToken());}
    public static long nextLong(){return Long.parseLong(inp.nextToken());}
    public static double nextDouble(){return Double.parseDouble(inp.nextToken());}
    
}