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.Scanner;


public class C121 {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long k = sc.nextInt();
        int[] a = new int[16];
        int min = Math.min(n, 15);
        int ans = -1;
        if (fac(min) >= k) {
            ArrayList<Integer> list1 = new ArrayList<Integer>();
            for (int i = min; i >= 1; i--) {
                long fact = fac(i - 1);
                int x = (int)(k / fact);
                if (k % fact != 0) {
                    x++;
                    k = k % fact;
                } else {
                    k = fact;
                }
                int h = 0, ind = 0;
                for (int j = 1; j <= min; j++) {
                    if (!list1.contains(j)) {
                        h++;
                    }
                    if (h == x) {
                        list1.add(j);
                        ind = j;
                        break;
                    }
                }
                a[min-i+1] = ind;
            }
            int[] b = new int[1024];
            int ind = 0;
            for (int i = 1; i <= 9; i++) {
                int pow = 1 << i;
                for (int j = 0; j < pow; j++) {
                    String s = Integer.toBinaryString(j);
                    while (s.length() < i)
                        s = "0"+s;
                    String ss = "";
                    for (int j1 = 0; j1 < i; j1++) {
                        if (s.charAt(j1) == '1')
                            ss += "7";
                        else
                            ss += "4";
                    }
                    b[++ind] = Integer.parseInt(ss);
                }
            }
            int count = 0;
            for (int i = 1; i <= ind; i++) {
                if (n - min < b[i]) {
                    count = i-1;
                    break;
                }
            }
            for (int i = 1; i <= min; i++) {
                if (Arrays.binarySearch(b, 1, ind+1, i+n-min) > 0 && Arrays.binarySearch(b, 1, ind+1, a[i]) > 0)
                    count++;
            }
            ans = count;
        }
        System.out.println(ans);
    }

    private static long fac(int n) {
        if (n <= 1)
            return 1;
        return n * fac(n-1);
    }

}
