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.Scanner;

public class Main {

	public static void main(String[] args) throws Exception {
		new Main().start();
	}

	private long fact(long m) {
		long ret = 1;
		
		for (long i = 1; i <= m; i++) {
			ret *= i;
		}
		return ret;
	}
	
	int[] data = new int[1022];
	long N, K;
	int[] ten = new int[10];
	
	private void start() {
		Scanner in = new Scanner(System.in);

		N = in.nextLong();
		K = in.nextLong();
		
		if (N < 13 && fact(N) < K) {
			System.out.println(-1);
			return;
		}
		
		ten[0] = 1;
		for (int i = 1; i < 10; i++) {
			ten[i] = ten[i - 1] * 10;
		}
		int c = 0;
		for (int i = 1; i < 10; i++) {
			int n = 1 << i;
			for (int j = 0; j < n; j++) {
				int tmp = 0;
				for (int k = 0; k < i; k++) {
					long q;
					if (0 != (j & (1 << k))) {
						q = 7;
					} else {
						q = 4;
					}
					tmp += q * ten[k];
				}
				data[c++] = tmp;
			}
		}
		
		Arrays.sort(data);
		
		long s[];
		
		if (N >= 13) {
			s = kth(13, K);
			
			for (int i = 0; i < 13; i++) {
				s[i] += N - 13;
			}
			int ans = 0;
			for (int i = 0; i < c && data[i] <= N; i++) {
				if (data[i] <= N - 13) {
					ans++;
				} else {
					if (is(s[(int) (data[i] - (N - 13))])) {
						ans++;
					}
				}
			}
			System.out.println(ans);
			
		} else {
			s = kth((int)N, K);
			
			int ans = 0;
			
			for (int i = 0; i < c && data[i] <= N; i++) {
				if (is(s[data[i]])) {
					ans++;
				}
			}
			System.out.println(ans);
		}
	}
	
	private boolean is(long v) {
		String s = Long.toString(v);
		
		for (int i = 0; i < s.length(); i++) {
			if (s.charAt(i) != '4' && s.charAt(i) != '7') {
				return false;
			}
		}
		return true;
	}
	
	private long[] kth(int n, long k) {
		ArrayList<Integer> c = new ArrayList<Integer>();
		for (int i = 1; i <= n; i++) {
			c.add(i);
		}
		long[] ret = new long[n + 1];
		for (int i = 1; i <= n; i++) {
			long t = fact(n - i);
			
			int ct = 1;
			while (k > t) {
				k -= t;
				ct++;
			}
			ret[i] = c.get(ct - 1);
			c.remove(ct - 1);
		}
		return ret;
	}

}
