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.*;
import java.util.*;

public class B implements Runnable {
	public static void main(String[] args) {
		new Thread(new B()).start();
	}

	BufferedReader br;
	StringTokenizer in;
	PrintWriter out;

	public String nextToken() throws IOException {
		while (in == null || !in.hasMoreTokens()) {
			in = new StringTokenizer(br.readLine());
		}

		return in.nextToken();
	}

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

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

	public double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	long[] chs;
	int n = 0;
	long pl, pr, vl, vr;
	long st;
	long en;

	void gen(long num) {
		if (num > en)
			return;
		if (num >= st)
			chs[n++] = num;
		gen(num * 10 + 7);
		gen(num * 10 + 4);
	}

	long inter(long pl2, long pr2, long l, long chs2) {
		long t = Math.min(pr2, chs2) - Math.max(pl2, l) + 1;
		if (t <= 0)
			return 0;
		return t;
	}

	public void solve() throws IOException {
		chs = new long[100000];
		pl = nextInt();
		pr = nextInt();
		vl = nextInt();
		vr = nextInt();
		int k = nextInt();
		st = Math.min(pl, vl);
		en = Math.max(pr, vr);
		gen(0);
		st--;
		Arrays.sort(chs, 0, n);
		chs[n] = en + 1;
		double res = 0;
		for (int i = 0; i + k - 1 < n; i++) {
			res += (double) inter(pl, pr, st + 1, chs[i])
					/ (double) (pr - pl + 1)
					* (double) inter(vl, vr, chs[i + k - 1], chs[i + k] - 1)
					/ (double) (vr - vl + 1);
			res += (double) inter(vl, vr, st + 1, chs[i])
					/ (double) (vr - vl + 1)
					* (double) inter(pl, pr, chs[i + k - 1], chs[i + k] - 1)
					/ (double) (pr - pl + 1);
			if ((k == 1) && (inter(pl, pr, chs[i], chs[i]) > 0)
					&& (inter(pl, pr, chs[i], chs[i]) > 0))
				res -= 1 / (double) (vr - vl + 1) / (pr - pl + 1);

			st = chs[i];
		}
		out.printf(Locale.US, "%.9f\n", res);

	}

	public void run() {
		try {
			Locale.setDefault(Locale.US);
			br = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);

			solve();

			out.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}
}