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.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

public class CLuckyPermutation {

	private static Map<Integer, Integer> map = new HashMap<>();

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);

		int n = scan.nextInt();
		int k = scan.nextInt() - 1;

		scan.close();
		
		if(k >= count(n)) {
			System.out.println(-1);
			return;
		}

		int[] factoradic = convertToFactoradic(k, n);

		List<Integer> last = new ArrayList<>();
		for (int i = Math.max(1, n - 12); i <= n; i++) {
			last.add(i);
		}

		int[] permutatedKth = new int[Math.min(n, 13)];

		int j = 0;
		for (int i = permutatedKth.length - 1; i >= 0; i--) {
			permutatedKth[j] = last.get(factoradic[i]);
			last.remove(factoradic[i]);
			j++;
		}

		createLuckyNumbers("");

		int res = 0;
		for (Integer i : map.keySet()) {
			if (i > n)
				;
			else if (i < Math.max(1, n - 12))
				res++;
			else if (map.containsKey(permutatedKth[(i <= permutatedKth.length ? i : 13 - (n - i)) - 1]))
				res++;
		}

		System.out.println(res);
	}
	
	static int count(int n) {
		if(n == 0)
			return 1;
		if(n > 12)
			return 2000000000;
		return count(n-1) * n;
	}

	static void createLuckyNumbers(String s) {
		if (s.length() >= 10)
			return;
		if (s.length() > 0) {
			int v = Integer.parseInt(s);
			map.put(v, 1);
		}
		createLuckyNumbers(s + "4");
		createLuckyNumbers(s + "7");
	}

	static int[] convertToFactoradic(int k, int n) {
		int[] res = new int[Math.min(n, 13)];

		int mod;
		for (int i = 1; i <= res.length; i++) {
			mod = k % i;
			k = k / i;
			res[i - 1] = mod;
		}

		return res;
	}

}