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.*;
import java.io.*;
public class C121 {
    public static long[] fact = new long[15];
    public static void main(String[] args) throws IOException {
        List<Long> list = new ArrayList<Long>();
        for (int len = 1; len <= 10; len++) {
            for (int which = 0; which < (1<<len); which++) {
                String str = "";
                long lucky = 0;
                long mult = 1;
                for (int bit = 0; bit < len; bit++) {
                    if ((which&(1<<bit))!=0) lucky += mult * 7;
                    else lucky += mult * 4;
                    mult *= 10;
                }
                list.add(lucky);
                if (len==10) break;
            }
        }
        fact[0] = 1;
        for (int i = 1; i < fact.length; i++) {
            fact[i] = ((long)i)*fact[i-1];
        }
        Scanner s = new Scanner(System.in);
        PrintStream out = System.out;
        long n = s.nextInt(), k = s.nextInt();
        if (n < fact.length && k > fact[(int)n]) {
            System.out.println("-1");
            return;
        }
        int digits = (int)Math.min(14, n);
        long[] lastDigits = new long[digits];
        long realIndex = n - digits + 1;
        for (int i = 0; i < digits; i++) {
            lastDigits[i] = realIndex + i;
        }
        lastDigits = applyPerm(lastDigits, (int)digits, (int)(k-1));
        long ct = 0;
        for (int i = 0; i < list.size(); i++) {
            if (list.get(i) <= n-digits) ct++;
            else break;
        }
        // ct += number of lucky in [1,n-digits]
        for (int i = 0; i < digits; i++) {
            if (lucky(realIndex+i) && lucky(lastDigits[i])) {
                ct++;
            }
        }
        //System.out.println(Arrays.toString(lastDigits));
        System.out.println(ct);
    }
    public static boolean lucky(long n) {
        String s = ""+n;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i)!='4' && s.charAt(i)!='7') return false;
        }
        return true;
    }
    public static long[] applyPerm(long[] array, int n, int k) {
        List<Integer> remain = new ArrayList<Integer>();
        int[] perm = new int[n];
        for (int i = 0; i < n; i++) remain.add(i);
        for (int at = 0; at < n; at++) {
            int i = n - 1 - at;
            int index = (int)(k/fact[i]);
            perm[at] = remain.get(index);
            k %= fact[i];
            remain.remove(index);
        }
        long[] newArray = new long[n];
        for (int i = 0; i < n; i++) {
            newArray[i] = array[perm[i]];
        }
        return newArray;
    }
}
