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.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;

/**
 * Created by IntelliJ IDEA.
 * User: serparamon
 * Date: 27.10.11
 * Time: 18:39
 * To change this template use File | Settings | File Templates.
 */
public class Contest20111027_C {

    private static Scanner input = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
    private static PrintWriter output = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    private static int max = 19;

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

    public static void main(String[] args) {
        int n = input.nextInt();
        int k = input.nextInt();
        long[] f = new long[max];
        f[0] = 1;
        for (int i=1; i<max; i++) {
            f[i] = i*f[i-1];
        }
        ArrayList<Long> s = new ArrayList<Long>();
        for (int i=1; i<11; i++) {
            for (int j=0; j<(1<<i); j++) {
                long x = 0;
                for (int q=0; q<i; q++) {
                    if ((j & (1<<q)) == 0) {
                        x = 10*x + 4;
                    }
                    else {
                        x = 10*x + 7;
                    }
                }
                s.add(x);
            }
        }
        Collections.sort(s);

        if (n<max && k>f[n]) {
            output.println(-1);
            output.flush();
            return;
        }

        int p = 0;
        while (k>f[p]) p++;
        int ans = 0;
        k--;
        boolean[] b = new boolean[p+1];
        Arrays.fill(b, false);
        for (int i=1; i<=p; i++) {
            int q = (int)(k/f[p-i])+1;
            int x=0;
            for (int j=1; j<=p; j++) {
                if (!b[j])
                    q--;
                if (q==0) {
                    b[j] = true;
                    x = j+n-p;
                    break;
                }
            }
            int pos = n-p+i;
            if (good(x) && good(pos)) {
                ans++;
            }
            k %= f[p-i];
        }
        for (long x : s) {
            if (x <= n-p)
                ans++;
        }
        output.println(ans);
        output.flush();
    }
}
