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.Scanner;


public class C121 {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        if (n < 13) {
            if (k > fac(n)) {
                System.out.println(-1);
                return;
            }
        }
        int[]a = new int[14];
        int m, ans = 0, add;
        if (n <= 13) {
            m = n;
            for (int i = 1; i <= m; i++) {
                a[i] = i;
            }
            add = 0;
        }
        else {
            m = 13;
            for (int i = 1; i <= 13; i++) {
                a[i] = n-13+i;
            }
            add = n-13;
            ans = num_lucly(n-13);
        }
        long[]fac = new long[m+1];
        fac[0] = 1;
        ArrayList<Integer> remain = new ArrayList<Integer>();
        for (int i = 1; i <= m; i++) {
            fac[i] = fac[i-1] * i;
            remain.add(i);
        }
        for (int i = 1; i <= m; i++) {
            int t = (int) (k / fac[m-i]);
            if (k % fac[m-i] != 0)
                t++;
            int ind = (t-1) % (m-i+1);
            int temp = remain.get(ind);
            remain.remove(ind);
            if (luckly(a[temp]) && luckly(i+add))
                ans++;
        }
        System.out.println(ans);
    }

    private static boolean luckly(int k) {
        String s = k+"";
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) != '4' && s.charAt(i) != '7')
                return false;
        }
        return true;
    }

    private static int num_lucly(int k) {
        int res = 0, L;
        String s;
        int lucly;
        for (int i = 1; i <= 9; i++) {
            if (i <= (k+"").length()) {
                for (int j = 0; j < (1 << i); j++) {
                    s = Integer.toBinaryString(j);
                    L = s.length();
                    for (int j2 = 1; j2 <= i-L; j2++) {
                        s = "0"+s;
                    }
                    lucly = 0;
                    for (int j2 = 0; j2 < i; j2++) {
                        lucly *= 10;
                        if (s.charAt(j2)=='0')
                            lucly += 4;
                        else
                            lucly += 7;
                    }
                    if (lucly <= k)
                        res++;
                }
            }
        }
        return res;
    }

    private static long fac(int n) {
        if (n < 2)
            return 1;
        return n * fac(n-1);
    }
}
