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 Main {
    public static long factorial(int n) {
        long res = 1;
        for (int i = n; i > 1; i--)
            res *= i;
        return res;
    }

    private static ArrayList<Integer> res(ArrayList<Integer> v, int k) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        long fact;
        int indx, limit = v.size();
        for (int i = 0; i < limit; i++) {
            fact = factorial(v.size() - 1);
            indx = (int) (k / fact);
            if (k % fact == 0)
                indx--;
            k -= fact * indx;
            int temp = v.get(indx);
            res.add(temp);
            v.remove(indx);
        }
        return res;
    }

    static int lucky[] = new int[1022];
    static int size = 0;

    private static void generate(int i, String s) {
        if (i == 10)
            return;
        if (s.length() > 0)
            lucky[size++] = new Integer(s);
        generate(i + 1, s + 4);
        generate(i + 1, s + 7);
    }

    public static void main(String[] args) {
        generate(0, "");
        Arrays.sort(lucky);
        Scanner myScanner = new Scanner(System.in);

        int k, n, startIndx = 0;
        n = myScanner.nextInt();
        k = myScanner.nextInt();
        for (int i = 0; i < 14; i++)
            if (factorial(i + 1) >= k) {
                startIndx = n - i;
                break;
            }
        if (startIndx <= 0) {
            System.out.println("-1");
            System.exit(0);
        }
        ArrayList<Integer> init = new ArrayList<Integer>();
        for (int i = startIndx; i <= n; i++)
            init.add(i);

        ArrayList<Integer> perm = res(init, k);

        int res = 0;
        for (int i = 0; i < perm.size(); i++)
            if (Arrays.binarySearch(lucky, i + startIndx) >= 0
                    && Arrays.binarySearch(lucky, perm.get(i)) >= 0)
                res++;

        for (int i = 0; i < lucky.length; i++)
            if (lucky[i] < startIndx)
                res++;
        
        System.out.println(res);
    }
}
