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.Arrays;
import java.util.Collections;
import java.util.Scanner;

public class B {
    static ArrayList<Long> numbers;
    static long max;
    static long min;

    public static void generate(String cur) {
        long numb = Long.parseLong(cur);
        if (numb > max) {
            return;
        }
        if (numb <= max && numb >= min)
            numbers.add(numb);
        generate(cur + '4');
        generate(cur + '7');
    }

    public static long getInterSection(long stIn1, long endIn1, long stIn2,
            long endIn2) {
        if (stIn2 > endIn1 || endIn2 < stIn1)
            return 0;
        long[] res = { stIn1, endIn1, stIn2, endIn2 };
        Arrays.sort(res);
        return res[2] - res[1] + 1;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long pl = sc.nextInt();
        long pr = sc.nextInt();
        long vl = sc.nextInt();
        long vr = sc.nextInt();
        int k = sc.nextInt();
        min = Math.min(vl, pl);
        max = Math.max(vr, pr);
        numbers = new ArrayList<Long>();
        generate("4");
        generate("7");
        Collections.sort(numbers);
        long total = (pr - pl+1) * (vr - vl + 1);
        long cur = 0;
        for (int i = 0; i + k - 1 < numbers.size(); i++) {
            long min = numbers.get(i);
            long max = numbers.get(i + k - 1);
            long nextMax, nextMin;
            if (i + k == numbers.size()) {
                nextMax = Integer.MAX_VALUE;
            } else {
                nextMax = numbers.get(i + k);
            }
            if (i == 0) {
                nextMin = 0;
            } else {
                nextMin = numbers.get(i - 1);
            }
            long left1 = getInterSection(nextMin+1, min , vl, vr);
            long right1 = getInterSection(max , nextMax-1, vl, vr);
            long left2 = getInterSection(nextMin+1, min , pl, pr);
            long right2 = getInterSection(max , nextMax-1, pl, pr);
            
            cur+=(left1*right2+left2*right1);
            if(min==max&&left1>0&&left2>0&&right1>0&&right2>0)
                cur-=1;
        }
        System.out.println((cur * 1.0) / total);

    }
}