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


public class C {
    static List<Integer> luckyNumbers = new ArrayList<Integer>();
    static long factorials[] = new long[13];
    
    static void calculateLuckyNumbers(int n , int i){
        if(i >= 10) return;
        
        int nn = n*10 + 4;
        luckyNumbers.add(nn);
        calculateLuckyNumbers(nn, i+1);
        
        nn = n*10 + 7;
        luckyNumbers.add(nn);
        calculateLuckyNumbers(nn, i+1);
        
    }
    static boolean isLucky(long x){
        while(x>0){
            int d =(int)( x%10);
            if(d != 4 && d!= 7 ){
                return false;
            }
            x/=10;
        }
        return true;
    }
    public static void main(String[] args) {
        calculateLuckyNumbers(0 , 1);
        factorials[1] = 1;
        
        for(int i = 2 ; i<=12 ; i++ ){
            factorials[i] = factorials[i-1]*(i);
        }
        Scanner in = new Scanner(System.in);
        long n = in.nextInt();
        long k = in.nextInt();
        in.close();
        if( n<13 && k>factorials[(int)n] ){
            System.out.println(-1);
        }else{
            long index = Math.max(0, n-13);
            int count = 0;
            
            for(Integer i : luckyNumbers){
                if(i<=index) count++;
            }
            
            List<Long> l = new LinkedList<Long>();
            List<Long> ll = new LinkedList<Long>();
            
            for( long i = index+1 ; i<=n ; i++){
                l.add(i);
            }
            k--;
            while(k>0){
                long r = k/factorials[l.size()-1];
                k = k%factorials[l.size()-1];
                long num = l.remove((int)r);
                ll.add(num);
            }
            ll.addAll(l);
            for(int i = 0 ; i<ll.size() ; i++){
                if(isLucky(i+index+1) && isLucky(ll.get(i))){
                    count++;
                }
            }
            
            System.out.println(count);
            
        }
    }

}
