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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class ProblemC {

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		
		String[] nk = in.readLine().split(" ");
		long n = Long.valueOf(nk[0]);
		long k = Long.valueOf(nk[1]);
		
		long perm = 1;
		int l = 0;
		while (perm < k) {
			l++;
			perm *= l;
		}
		if (l > n) {
			out.println(-1);
			out.flush();
			return;
		}

		out.println(solveA(n - l) + solveB(n - l + 1, n, k-1));
			
		

		out.flush();
	}
	
	
	private static long solveB(long f, long t, long k) {
		if (f > t) {
			return 0;
		}
		long luckluck = 0;
		
		long d = (t - f) + 1;
		List<Long> remain = new ArrayList<Long>();
		for (long x = f ; x <= t ; x++) {
			remain.add(x);
		}
		long index = f;
		while (index <= t) {
			long perm = 1;
			for (long p = 1 ; p < d ; p++) {
				perm *= p;
			}
			int canHave = (int)(k / perm);
			k -= canHave * perm;
			long selected = remain.remove(canHave);
			if (islucky(index) && islucky(selected)) {
				luckluck++;
			}
			index++;
			d--;
		}
		return luckluck;
	}

	static boolean islucky(long num) {
		return String.valueOf(num).replaceAll("4", "").replaceAll("7", "").length() == 0;
	}
	

	static List<Long> lucky;
	
	public static int dfs(long max, long x) {
		if (x > max) {
			return 0;
		}
		return 1 + dfs(max, x*10+4) + dfs(max, x*10+7);
	}
	
	private static int solveA(long l) {
		return dfs(l, 4) + dfs(l, 7);
	}

	public static void debug(Object... o) {
		System.err.println(Arrays.deepToString(o));
	}
}
