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.Scanner;

public class e {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt(), k = in.nextInt();
		if(k > count(n)) {
			System.out.println(-1);
			return;
		}
		int ans = 0;
		int mi = Math.max(1, n-13);
		find("");
		for(long l : list) {
			if(l < mi)
				ans++;
		}
		ArrayList<Integer> unused = new ArrayList<>();
		
		for(int i=mi;i<=n;i++) {
			unused.add(i);
		}
		
		ans += solve(mi, unused, k);
		System.out.println(ans);
	}
	static int count(int n) {
		if(n == 0)
			return 1;
		if(n > 12)
			return 2000000000;
		return count(n-1) * n;
	}
	static boolean isLucky(int v) {
		String s = v + "";
		boolean ans = true;
		for(int i=0;i<s.length();i++) {
			ans &= s.charAt(i) == '4' || s.charAt(i) == '7';
		}
		return ans;
	}
	static int solve(int idx, ArrayList<Integer> unused, int k) {
		if(unused.size() == 0)
			return 0;
		int ncnt = count(unused.size()-1);
		for(int i=0;i<unused.size();i++) {
			if(k <= ncnt) {
//				System.out.print(unused.get(i) + "[" + idx + "]" + " ");
				boolean lucky = isLucky(idx) && isLucky(unused.get(i));
				unused.remove(i);
				return solve(idx+1, unused, k) + (lucky ? 1 : 0);
			}
			k -= ncnt;
		}
		return 0;
	}
	static ArrayList<Long> list = new ArrayList<>();
	static void find(String s) {
		if(s.length() >= 11)
			return;
		if(s.length() > 0) {
			long v = Long.parseLong(s);
			list.add(v);
		}
		find(s + "4");
		find(s + "7");
	}
	
}
