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.*;

public class Main {

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

  int solve(int n, int k) {

    int last = 1;
    long perm = 1;

    while( k > perm ) {
      ++last;
      perm *= last;
    }

    if( last > n ) { return - 1; }

    int ret = count(n - last);
    int ans[] = createKth(last, k - 1);
    for(int i=0; i<last; ++i) {
      int pos = i + n - last + 1;
      int val = ans[i] + n - last + 1;
      if( ok(pos) && ok(val) ) { ++ret; }
    }
    return ret;

  }

  int[] createKth(int digs, int k) {
    int ret[] = new int[digs];
    for(int i=0; i<digs; ++i) { ret[i] = i; }
    int count = 0;
    long use = 1;
    while( count < digs - 1 ) {
      ++count;
      use *= count;
    }
    for(int i=0; i<digs-1; ++i) {
      int find = (int)(k / use);
      int set = ret[i + find];
      for(int j=find; j>0; --j) {
        ret[i + j] = ret[i + j - 1];
      }
      ret[i] = set;
      k = (int)(k % use);
      use /= count;
      --count;
    }
    return ret;
  }

  boolean ok(int n) {
    while( n > 0 ) {
      int k = n % 10;
      if( !(k == 4 || k == 7) ) { return false; }
      n /= 10;
    }
    return true;
  }

  int count(int n) {
    int ret = 0;
    for(int i=1; i<12; ++i) {
      int m = 1 << i;
      for(int j=0; j<m; ++j) {
        long cur = 0;
        long base = 1;
        for(int k=0; k<i; ++k) {
          if( (j & (1 << k)) == 0 ) { cur += base * 4; }
          else { cur += base * 7; }
          base *= 10;
        }
        if( cur <= n ) { ++ret; }
      }
    }
    return ret;
  }

  void solve() {

    Scanner cin = new Scanner(System.in);

    while( cin.hasNextInt() ) {

      int n = cin.nextInt();
      int k = cin.nextInt();

      System.out.println(solve(n, k));

    }

  }

}
