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.util.LinkedList;
import java.util.StringTokenizer;
import java.util.TreeSet;


public class LuckyPermutation {
	
	static class Scanner{
		BufferedReader br=null;
		StringTokenizer tk=null;
		public Scanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		public String next() throws IOException{
			while(tk==null || !tk.hasMoreTokens())
				tk=new StringTokenizer(br.readLine());
			return tk.nextToken();
		}
		public int nextInt() throws NumberFormatException, IOException{
			return Integer.valueOf(next());
		}
	}
	
	static long[] fact = new long[14];
	static TreeSet<Long> set = new TreeSet<Long>();
	
	static void init(){
		fact[0] = 1;
		for(int i = 1; i <= 13; i++)
			fact[i] = fact[i - 1]*i;
		for(int tam = 1; tam <= 11; tam++){
			for(int i = 0;  i < (1<<tam); i++){
				long num = 0;
				for(int k = tam - 1; k >= 0; k--){
					int n= ( (((i>>k) & 1) == 1)? 7: 4);
					num *= 10;
					num += n;
				}
				set.add(num);
			}
		}
		return;
	}
	
	public static void main(String args[]) throws NumberFormatException, IOException{
		Scanner sc = new Scanner();
		init();
		int N = sc.nextInt();
		int K = sc.nextInt();
		LinkedList<Integer> ans = new LinkedList<Integer>();
		if (N <= 13 && fact[N] < K){
			System.out.println("-1");
			return;
		}
		int amount = 0;
		long left = 1;
		long right = N - 13;
		for(long lucky: set){
			if (lucky>= left && lucky<=right)
				amount++;
		}
		TreeSet<Integer> current = new TreeSet<Integer>();
		for(int i = Math.max(1, N - 12); i <= N; i++)
			current.add(i);
		int limit = current.size();
		for(int i = 0; i < limit; i++){
			int index = 1;
			for(int e: current){
				if (fact[current.size() - 1]*index >= K){
					K -= fact[current.size() - 1] * (index - 1);
					current.remove(e);
					ans.add(e);
					break;
				}
				index++;
			}
		}
		int pos = Math.max(N - 12,1);
		for(int e: ans){
			if (set.contains(pos + 0L) && set.contains(e + 0L))
				amount++;
			pos++;
		}
		System.out.println(amount);
	}

}
