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.*;

public class cf121c {
    static TreeSet<Long> lucky = new TreeSet<Long>();
    static long[] facts = new long[20];
    public static void main(String[] args) {
        genLucky(0,12);
        facts[0] = facts[1] = 1;
        for(int i=2; i<20; i++)
            facts[i] = facts[i-1]*i;
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        if(n < 20 && facts[n] < k) {
            System.out.println(-1);
            return;
        }
            
        int shift = n-20;
        long[] v = new long[20];
        for(int i=19; i>=0; i--)
            v[i] = n-(19-i);
        int ret = lucky.headSet((long)shift, true).size();
        
        k--;
        long[] last = new long[20];
        for(int i=0; i<20; i++) {
            last[i] = k/facts[19-i];
            k %= facts[19-i];
        }
        int[] lasti = new int[20];
        ArrayList<Integer> tmp = new ArrayList<Integer>();
        for(int i=0; i<20; i++)
            tmp.add(i);
        for(int i=0; i<20; i++)
            lasti[i] = tmp.remove((int)last[i]);
        long[] lastv = new long[20];
        for(int i=0; i<20 ;i++)
            lastv[i] = v[lasti[i]];
        
        for(int i=0; i<20; i++) 
            if(lucky.contains(lastv[i]) && lucky.contains((long)(shift+i+1)))
                ret++;
        
        System.out.println(ret);
    }
    static void genLucky(long cur, int left) {
        if(left == 0) return;
        if(cur != 0) lucky.add(cur);
        genLucky(cur*10+7, left-1);
        genLucky(cur*10+4, left-1);
    }
}

/*

7 4

4 7


*/
