Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

One day Petya dreamt of a lexicographically k-th permutation of integers from 1 to n. Determine how many lucky numbers in the permutation are located on the positions whose indexes are also lucky numbers.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 109) — the number of elements in the permutation and the lexicographical number of the permutation.

Output

If the k-th permutation of numbers from 1 to n does not exist, print the single number "-1" (without the quotes). Otherwise, print the answer to the problem: the number of such indexes i, that i and ai are both lucky numbers.

Examples

Input

7 4


Output

1


Input

4 7


Output

1

Note

A permutation is an ordered set of n elements, where each integer from 1 to n occurs exactly once. The element of permutation in position with index i is denoted as ai (1 ≤ i ≤ n). Permutation a is lexicographically smaller that permutation b if there is such a i (1 ≤ i ≤ n), that ai < bi, and for any j (1 ≤ j < i) aj = bj. Let's make a list of all possible permutations of n elements and sort it in the order of lexicographical increasing. Then the lexicographically k-th permutation is the k-th element of this list of permutations.

In the first sample the permutation looks like that:

1 2 3 4 6 7 5

The only suitable position is 4.

In the second sample the permutation looks like that:

2 1 3 4

The only suitable position is 4.
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF121C extends PrintWriter {
	CF121C() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF121C o = new CF121C(); o.main(); o.flush();
	}

	long lucky(int b) {
		return b == 1 ? 0 : lucky(b / 2) * 10 + (b % 2 == 0 ? 4 : 7);
	}
	boolean is_lucky(int y) {
		return y == 0 ? true : is_lucky(y / 10) && (y % 10 == 4 || y % 10 == 7);
	}
	static final int N = 13;
	boolean[] used = new boolean[N];
	int[] ff = new int[N];
	void init() {
		int f = 1;
		for (int i = 0; i < N; i++) {
			ff[i] = f;
			f *= i + 1;
		}
	}
	int get(int n, int k, int i) {
		Arrays.fill(used, false);
		for (int h = 0; h < n; h++) {
			int f = ff[n - 1 - h];
			int s = k / f;
			for (int j = 0; j < n; j++)
				if (!used[j] && s-- == 0) {
					if (h == i)
						return j;
					used[j] = true;
					break;
				}
			k %= f;
		}
		return -1;
	}
	void main() {
		init();
		int n = sc.nextInt();
		int k = sc.nextInt();
		long f = 1;
		int n_ = 1;
		while (f < k)
			f *= ++n_;
		if (n_ > n) {
			println(-1);
			return;
		}
		int ans = 0;
		for (int b = 2; ; b++) {
			long x = lucky(b);
			if (x > n)
				break;
			if (x <= n - n_) {
				ans++;
				continue;
			}
			int y = n - n_ + get(n_, k - 1, (int) (x - (n - n_) - 1)) + 1;
			if (is_lucky(y))
				ans++;
		}
		println(ans);
	}
}
