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.* ;
import java.util.*;
import static java.lang.Math.* ;
import static java.util.Arrays.* ;

public class Main {
	
	public static void main(String[] args) {
		
			new Main().solveProblem();
		
		out.close();
	}

	static Scanner in = new Scanner(new InputStreamReader(System.in));
	static PrintStream out = new PrintStream(new BufferedOutputStream(System.out));
	
	long[] fac = new long[21] ;
	public void solveProblem() {	
		
		fac[0] = 1 ;
		for( int i = 1 ; i <= 20 ; i++ )
			fac[i] = fac[i-1]*i ;

		int n = in.nextInt() ;
		int k = in.nextInt() ;
		
		if( n <= 20 && fac[n] < k ){
			System.out.println(-1);
			return ;
		}
		
		int t = 0 ;
		while( fac[t] < k )
			t++ ;

		
	//System.out.println( n - t + 1 );
		int a = luckyNumberUnder( n - t + 1 ) ;
		int b = ("" + (n - t + 1)).length() - 1 ;
		for( int i = 1 ; i <= b ; i++ )
			a += (1<<i ) ;

		int[] perm = new int[t] ;
		for( int i = 0 ; i < t ; i++ )
			perm[i] = n - t + 1 + i ;
		
		int[] nperm = berekenPerm( k-1, 0, perm ) ;
		for( int i = 0 ; i < t ; i++ )
			if( isLucky(n-t+1+i) && isLucky(nperm[i]))
				a++ ;
		//System.out.println(Arrays.toString(nperm));
		System.out.println(a);

	
	}
	
	private int[] berekenPerm(int k, int i, int[] perm) {
		
		int n = perm.length ;
		
		if( i == n )
			return perm ;
		
		sort( perm, i, n) ;
		int d = (int) (k * ( n - i ) / fac[n-i]) ;
		//System.out.println((k*(n-i)) + " " + fac[n-i]);
		
		int tmp = perm[i+d] ;
		perm[i+d] = perm[i] ;
		perm[i] = tmp ;

		return berekenPerm( (int) (k % fac[n-i-1]), i+1, perm );
	}

	private int luckyNumberUnder(int i) {
		
		String s = i + "" ;
		
		int a = 0 ;
		if( s.charAt(0) > '4')
			a += (1 << s.length()-1) ;
		if( s.charAt(0) > '7' )
			a += (1 << s.length()-1) ;
		if( s.length() > 1 && (s.charAt(0)  == '7' || s.charAt(0) == '4' ))
			a += luckyNumberUnder( Integer.parseInt(s.substring(1))) ;
		
		return a;
	}

	boolean isLucky( int get ){
	
		while( get > 0 ){
			if( get % 10 != 4 && get % 10 != 7 )
				return false ;
			get /= 10 ;
		}
		
		return true ;
	}
       
}