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 static java.lang.Math.*;
import static java.util.Arrays.*;
import java.util.*;

public class Main {
    public static void main(String[]args){
        new Main().run();
    }
    
    private void debug(Object...os){
        System.err.println(deepToString(os));
    }

    private void run() {
        Scanner sc = new Scanner(System.in);
        int n=sc.nextInt(),K=sc.nextInt();
        long[] fact = new long[19];
        for (int i = 0; i < 19; i++)fact[i] = i==0 ? 1 : fact[i-1]*i;
        int tail = min(n,18);
        if(fact[tail] < K){
            System.out.println(-1);return;
        }
        int off = n-tail+1;
        int res = 0;
        boolean[] used = new boolean[tail];
        for (int i = 0; i < tail; i++){
            for (int j = 0; j < tail; j++)if(!used[j]){
                if(fact[tail-1-i] >= K){
                    used[j] = true;
                    if(lucky(off+j) && lucky(off+i))res++;
                    break;
                }else{
                    K -= fact[tail-1-i];
                }
            }
        }
        for (int i = 1; i < 11; i++){
            for (int j = 0; j < 1<<i; j++){
                long val=0;
                for (int k = 0; k < i; k++){
                    val*=10;
                    val+= ((j>>k&1)==1) ? 4:7;
                }
                if(val <= n-tail)res++;
            }
        }
        System.out.println(res);
    }

    private boolean lucky(int i) {
        if(i==0)return false;
        while(i>0){
            if(i%10!=7 && i%10!=4)return false;
            i/=10;
        }
        return true;
    }
}
