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.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Locale;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.Map.Entry;

public class Solution implements Runnable {

	public static void main(String[] args) {
		(new Thread(new Solution())).start();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer st;

	String nextToken() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			String r = in.readLine();
			if (r == null) return null;
			st = new StringTokenizer(r);
		}
		return st.nextToken();
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
	
	long per(long a, long b, long c, long d) {
		if (a <= c) {
			if (b < c) return 0;
			return Math.min(b, d) - c + 1;
		} else {
			if (d < a) return 0;
			return Math.min(b, d) - a + 1;
		}
	}
	
	void solve() throws Exception {
		int pl = nextInt();
		int pr = nextInt();
		int vl = nextInt();
		int vr = nextInt();
		int k = nextInt();
		long[] a = new long[(1 << 15) + 1];
		for (int i = 0; i < (1 << 15); i++) {
			a[i + 1] = 0;
			int j = i + 2;
			long st = 1;
			while (j / 2 > 0) {
				a[i + 1] += st * ((j % 2 == 0)? 4 : 7);
				st *= 10; j /= 2;
			}
		}
		long ans = 0;
		for (int i = 1; i < (1 << 15); i++) {
			int j = i + k - 1;
			if (j >= (1 << 15)) break;
			ans += per(a[i - 1] + 1, a[i], pl, pr) * per(a[j], a[j + 1] - 1, vl, vr);
			ans += per(a[i - 1] + 1, a[i], vl, vr) * per(a[j], a[j + 1] - 1, pl, pr);
			if (i == j && pl <= a[i] && a[i] <= pr && vl <= a[i] && a[i] <= vr) ans--;
		}
		out.printf("%.10f", (double)ans / (double)((long)(pr - pl + 1) * (long)(vr - vl + 1)));
	}

	public void run() {
		Locale.setDefault(Locale.UK);
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
//			in = new BufferedReader(new FileReader("input.txt"));
//			out = new PrintWriter("output.txt");
//			in = new BufferedReader(new FileReader("granama.in"));
//			out = new PrintWriter(".out");
			solve();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		} finally {
			out.flush();
		}
	}

}