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.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class D {
    public static long get(long[] A, int k, long pl, long pr, long vl, long vr) {
        long ways = 0;
        for (int i = 0; i + k <= 1022; i++) {
            long lp = i == 0 ? pl : Math.max(pl, A[i - 1] + 1);
            long rp = Math.min(pr, A[i]);
            long left = Math.max(0, rp - lp + 1);
            long lv = Math.max(vl, A[i + k - 1]);
            long rv = i + k == 1022 ? vr : Math.min(vr, A[i + k] - 1);
            long right = Math.max(0, rv - lv + 1);
            ways += left * right;
        }
        return ways;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long pl = in.nextLong();
        long pr = in.nextLong();
        long vl = in.nextLong();
        long vr = in.nextLong();
        int k = in.nextInt();
        int tot = 1022;
        long[] A = new long[tot];
        int index = 0;
        Queue<Long> L = new LinkedList<Long>();
        L.add(4L);
        L.add(7L);
        while (index < tot) {
            long x = L.poll();
            A[index++] = x;
            L.add(x * 10 + 4);
            L.add(x * 10 + 7);
        }
        long ways = 0;
        ways += get(A, k, pl, pr, vl, vr);
        ways += get(A, k, vl, vr, pl, pr);
        if (k == 1)
            for (int i = 0; i < tot; i++)
                if (A[i] >= pl && A[i] <= pr && A[i] >= vl && A[i] <= vr)
                    ways--;
        long p = pr - pl + 1;
        long v = vr - vl + 1;
        System.out.println(ways * 1.0 / (p * v));
    }
}