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 Main {

        public static void main(String[] args) throws Exception {
                new Main().start();
        }

        private long fact(long m) {
                long ret = 1;
                
                for (long i = 1; i <= m; i++) {
                        ret *= i;
                }
                return ret;
        }
        
        int[] data = new int[1022];
        long N, K;
        int[] ten = new int[10];
        
        private void start() {
                Scanner in = new Scanner(System.in);

                N = in.nextLong();
                K = in.nextLong();
                
                if (N < 13 && fact(N) < K) {
                        System.out.println(-1);
                        return;
                }
                
                ten[0] = 1;
                for (int i = 1; i < 10; i++) {
                        ten[i] = ten[i - 1] * 10;
                }
                int c = 0;
                for (int i = 1; i < 10; i++) {
                        int n = 1 << i;
                        for (int j = 0; j < n; j++) {
                                int tmp = 0;
                                for (int k = 0; k < i; k++) {
                                        long q;
                                        if (0 != (j & (1 << k))) {
                                                q = 7;
                                        } else {
                                                q = 4;
                                        }
                                        tmp += q * ten[k];
                                }
                                data[c++] = tmp;
                        }
                }
                
                Arrays.sort(data);
                
                long s[];
                
                if (N >= 13) {
                        s = kth(13, K);
                        
                        for (int i = 0; i < s.length; i++) {
                                s[i] += N - 13;
                        }
                        int ans = 0;
                        for (int i = 0; i < c && data[i] <= N; i++) {
                                if (data[i] <= N - 13) {
                                        ans++;
                                } else {
                                        if (is(s[(int) (data[i] - (N - 13))])) {
                                                ans++;
                                        }
                                }
                        }
                        System.out.println(ans);
                        
                } else {
                        s = kth((int)N, K);
                        
                        int ans = 0;
                        
                        for (int i = 0; i < c && data[i] <= N; i++) {
                                if (is(s[data[i]])) {
                                        ans++;
                                }
                        }
                        System.out.println(ans);
                }
        }
        
        private boolean is(long v) {
                String s = Long.toString(v);
                
                for (int i = 0; i < s.length(); i++) {
                        if (s.charAt(i) != '4' && s.charAt(i) != '7') {
                                return false;
                        }
                }
                return true;
        }
        
        private long[] kth(int n, long k) {
                ArrayList<Integer> c = new ArrayList<Integer>();
                for (int i = 1; i <= n; i++) {
                        c.add(i);
                }
                long[] ret = new long[n + 1];
                for (int i = 1; i <= n; i++) {
                        long t = fact(n - i);
                        
                        int ct = 1;
                        while (k > t) {
                                k -= t;
                                ct++;
                        }
                        ret[i] = c.get(ct - 1);
                        c.remove(ct - 1);
                }
                return ret;
        }

}