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.Collections;
import java.util.Scanner;

public class Main {
    static ArrayList<Long> aa;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt(),k = in.nextInt();
        aa = new ArrayList<Long>(2500);
        dfs(7);dfs(4);
        int nn = n,l = Math.min(13, n);
        n = l;
        if (n < 13 && k > fact(n)) {
            System.out.println(-1);
            System.exit(0);
        }
        int[] a = new int[l];
        boolean[] v = new boolean[l];
        for (int i = 0; i < a.length; i++)
            a[i] = nn - i;
        Arrays.sort(a);
        int[] ans = new int[a.length];
        long g, kk = k, f;
        boolean d;
        for (int i = 0; i < a.length; i++) {
            g = kk / (f = fact(n - 1)) + 1;
            if (kk % f == 0)
                g--;
            int cnt = 0, j;
            for (j = 0; j < a.length; j++) {
                if (!v[j])
                    cnt++;
                if (cnt == g)
                    break;
            }
            v[j] = true;
            ans[i] = a[j];
            d = false;
            if (kk >= f)
                d = true;
            kk %= f;
            if (d && kk == 0)
                kk += f;
            n--;
        }
        Collections.sort(aa);
        int ret = 0;
        for (int i = 0; i < aa.size(); i++)
            if (aa.get(i) > nn - 13 + 1) {
                ret = i;
                break;
            }
        for (int i = 0; i < ans.length; i++)
            if (aa.contains((long) ans[i])
                    && aa.contains((long) (i + nn - l + 1)))
                ret++;
        System.out.println(ret);
    }
    public static long fact(long n) {
        if (n == 1 || n == 0)
            return 1;
        return n * fact(n - 1);
    }
    public static void dfs(long n) {
        aa.add(n);
        if (n > 1000000000)
            return;
        dfs(n * 10 + 4);
        dfs(n * 10 + 7);
    }
}