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.*;
import java.io.*;

public class Main {
	public static void main ( String args[] ) throws Exception {
		(new Task23()).solve();
	}
}

class Task23
{
	Scanner in;
	
	int[] getPermutation ( int[] a, long k )
	{
		boolean seen[] = new boolean[a.length];
		int r[] = new int[a.length];
		Arrays.fill ( r, -1 );
		for ( int i = 0; i < a.length; ++i )
		{
			long fact = 1;
			for ( int j = 1; j < a.length-i; ++j )
				fact *= j;
			long id = (k-1) / fact;
			k -= id*fact;
			for ( int j = 0; j < a.length; ++j )
				if ( !seen[j] )
				{
					if ( id == 0 ) {
						r[i] = a[j];
						seen[j] = true;
						break;
					}
					id--;
				}
		}
		return r;
	}
	
	boolean isLucky ( int n ) {
		for ( ; n > 0; n /= 10 )
			if ( n%10 != 4 && n%10 != 7 )
				return false;
		return true;
	}
	
	int countLucky ( long num, long limit ) {
		if ( num > limit )
			return 0;
		int ans = 1;
		ans += countLucky ( num*10 + 4, limit );
		ans += countLucky ( num*10 + 7, limit );
		return ans;
	}
	
	public void solve ( ) throws Exception
	{
		in = new Scanner ( System.in );
		int n = in.nextInt();
		long k = in.nextLong();
		
		if ( n <= 15 )
		{
			long fact = 1;
			for ( int i = 1; i <= n; ++i )
				fact *= i;
			if ( k > fact )
			{
				println ( "-1\n" );
				return;
			}
		}
		
		int last[] = new int[Math.min(15,n)];
		for ( int i = 0; i < last.length; ++i )
			last[i] = n-last.length+i+1;
		
		int ans = 0;
		last = getPermutation ( last, k );
		for ( int i = 0; i < last.length; ++i )
			if ( isLucky(last[i]) && isLucky(n-last.length+i+1) )
				ans++;
		
		ans += countLucky ( 0, n-last.length )-1;
		println ( ans );
	}
	
	public static void println ( Object o ) { System.out.println ( o ); }
	public static void print ( Object o ) { System.out.print ( o ); }
	public static void println ( ) { System.out.println ( ); }
}
