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.LinkedList;
import java.util.List;
import java.util.Scanner;

public class luckPerm {
    public static void debug(Object... obs) {
        System.out.println(Arrays.deepToString(obs));
    }
    

    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        
        long n=sc.nextLong();
        long k=sc.nextLong();
        
        List<Long>q=new ArrayList<Long>();
        int idx=0;
        long l=0;
        while(l <= n)
        {
            long l1=l*10 + 4;
            long l2=l*10 + 7;
            if(l1 > n) break;
            q.add(l1);
            if(l2 > n) break;
            q.add(l2);
            l = q.get(idx);
            idx++;
        }
        int cnt=0;
        
        //debug(q);
        
        if(n <= 13 && fac(n) < k)
        {
            System.out.println(-1);
            return;
        }
        
        //int[]mapit=new int[(int)n];
        //for(int v=1;v<=n;v++)
        for(long v : q)
        {
            //debug(v,n,k);
            long id=pos(v,n,k);
            //debug(v,n,k,id);
            //mapit[(int)id-1]=v;
            if(id == v)
                cnt++;
        }
        //debug(mapit);
        
        System.out.println(cnt);
    }
    private static long pos(long v, long n, long k) {
        
        long res=0;
        //debug(v,n,k);
        
        if(n > 14)
        {
            long dif=Math.min(n-14, v-1);
            res+=dif;
            n-=dif;
            v-=dif;
        }
        if(n==1)
        {
            res+=1;
            return res;
        }
        
        if(v==1 && n > 14)
        {
            return res+=1;
        }
//      if(n==2 && v==2)
//      {
//          if(k%2==0) res+=2;
//          else res+=1;
//          return res;
//      }
//      if(n==2 && v==1)
//      {
//          if(k%2==0) res+=1;
//          else res+=2;
//          return res;
//      }
        //debug(v,n,k);
        long fc=fac(n-1);
        long nk = (k%fc==0)?fc:k%fc;
        
        if( k <= (v-1)*fc)
        {
            
            res+= 1 + pos(v-1,n-1,nk);
        }
        else if( k <= v*fc)
        {
            res+=1;
        }
        else if(k <= n*fc)
        {
            res+= 1 + pos(v,n-1,nk);
        }
        else
        {
            return -1;
        }
        
        return res;
    }

    static long fac(long a)
    {
        long res=1;
        for(long i=1;i<=a;i++)
        {
            res*=i;
        }
        return res;
    }
}
