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.HashSet;
import java.util.Scanner;

public class E {
    static int[] Fac;
    static int[] P;
    static int n;
    static HashSet<Integer> H = new HashSet<Integer>();
    static int count = 0;
    static int start;
    static ArrayList<Long> Q = new ArrayList<Long>();
    static int inf = (int) 1e9;

    public static boolean isLucky(int x) {
        while (x > 0) {
            if (x % 10 != 4 && x % 10 != 7)
                return false;
            x /= 10;
        }
        return true;
    }

    public static void generate(int pos, int k) {
        if (pos > n)
            return;
        for (int i = start; i <= n; i++) {
            if (H.contains(i))
                continue;
            int left = n - start - H.size();
            if (Fac[left] < k) {
                k -= Fac[left];
            } else {
                H.add(i);
                if (isLucky(pos) && isLucky(i))
                    count++;
                generate(pos + 1, k);
            }
        }
    }

    public static void generate(long x) {
        Q.add(10 * x + 4);
        Q.add(10 * x + 7);
        if (x < inf) {
            generate(10 * x + 4);
            generate(10 * x + 7);
        }
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        int k = in.nextInt();
        Fac = new int[13];
        Fac[0] = 1;
        for (int i = 1; i < 13; i++)
            Fac[i] = Fac[i - 1] * i;
        if (n < 13 && k > Fac[n]) {
            System.out.println(-1);
            return;
        }
        start = 12;
        while (start >= 0 && Fac[start] >= k)
            start--;
        int pos = start = n - start;
        generate(pos, k);
        generate(0);
        for (long x : Q)
            if (x < start)
                count++;
        System.out.println(count);
    }
}
