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 java.lang.reflect.Array;
import java.math.*;
public class lucksum {
	public static long go(long m)
	{
		long res=0;
		Queue<Long> l=new LinkedList<Long>();
		l.add(new Long(0));
		do{
			Long t=l.remove();
			if(t.compareTo(new Long(m))<=0)
			{
				res++;
				l.add(t*10+4);
				l.add(t*10+7);
				}
			else
				break;
				
		}while(true);
		
		return res-1;
	}
	public static boolean isluck(long m)
	{	
		boolean res=true;
		while(m!=0)
		{
			long t=m%10;
			if((t!=4)&&(t!=7))
				return false;
			m/=10;
		}
		return res;
		
	}
	public static void main(String[] args)
	{

		//try{
			Scanner s=new Scanner(System.in);
			long n=s.nextLong();
			long k=s.nextLong();
			long []p=new long[21];
			Arrays.fill(p,Long.MAX_VALUE);
			long start=1;
			long mul=2;
			p[0]=1;
			while(start<=k)
			{
				p[(int)mul-1]=start;
				start*=mul;
				mul++;
			}
			if((n<20)&&(p[(int)n]<k))
			{
				System.out.println(-1);
				return;
			}
			int index=0;
			for(;p[index]<k;index++);
	//		index++;
			long res=go(n-index);

			List<Long> left=new LinkedList<Long>();
			for(int i=index-1;i>=0;i--)
				left.add(n-i);
			
			while(index>0)
			{
				Long[] mm=(Long[])left.toArray(new Long[left.size()]);
				
				long t=(k-1)/p[index-1];
				long h=mm[(int)t];
				if(isluck(h)&&isluck(n-index+1))
					res++;
				left.remove((int)t);
				k-=t*p[index-1];
				index--;

			}
			System.out.println(res);
	//	}catch (Exception e1){
			
		//}

	}
}


