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

public class Main {
	static ArrayList<Long> aa;

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		aa = new ArrayList<Long>(2500);
		dfs(7);
		dfs(4);
		int nn = n;
		int l = Math.min(13, n);
		n = l;
		if (n < 13 && k > fact(n)) {
			System.out.println(-1);
			System.exit(0);
		}
		int[] a = new int[l];
		boolean[] v = new boolean[l];
		for (int i = 0; i < a.length; i++)
			a[i] = nn - i;
		Arrays.sort(a);
		int[] ans = new int[a.length];
		long g;
		long kk = k;
		long f;
		boolean d;
		for (int i = 0; i < a.length; i++) {
			g = kk / (f = fact(n - 1)) + 1;
			if (kk % f == 0)
				g--;
			int cnt = 0, j;
			for (j = 0; j < a.length; j++) {
				if (!v[j])
					cnt++;
				if (cnt == g)
					break;
			}
			v[j] = true;
			ans[i] = a[j];
			d = false;
			if (kk >= f)
				d = true;
			kk %= f;
			if (d && kk == 0)
				kk += f;
			n--;
		}
		Collections.sort(aa);
		int ret = 0;
		for (int i = 0; i < aa.size(); i++)
			if (aa.get(i) > nn - 13 + 1) {
				ret = i;
				break;
			}
		for (int i = 0; i < ans.length; i++)
			if (aa.contains((long) ans[i])
					&& aa.contains((long) (i + nn - l + 1)))
				ret++;
		System.out.println(ret);
	}

	public static long fact(long n) {
		if (n == 1 || n == 0)
			return 1;
		return n * fact(n - 1);
	}

	public static void dfs(long n) {
		aa.add(n);
		if (n > 1000000000)
			return;
		dfs(n * 10 + 4);
		dfs(n * 10 + 7);
	}
}