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.
#include <bits/stdc++.h>
int ans, n, k, m;
bool its[99];
void count(long long l) {
  if (l <= n - m) ++ans, count(l * 10 + 4), count(l * 10 + 7);
}
bool isLucky(int n) {
  if (n == 0) return 0;
  while (n) {
    if (n % 10 != 4 && n % 10 != 7) return false;
    n /= 10;
  }
  return true;
}
int main() {
  int fac[15];
  fac[0] = 1;
  for (int i = 1; i <= 13; ++i) fac[i] = fac[i - 1] * i;
  scanf("%d %d", &n, &k);
  m = 1;
  while (fac[m] < k && m <= n) ++m;
  if (m > 13 || m > n)
    puts("-1");
  else {
    count(4), count(7);
    for (int i = 1; i <= m; ++i) {
      int x = 0, y = 1;
      while (k > fac[m - i]) k -= fac[m - i], ++y;
      for (int j = 1; j <= y; ++j) {
        ++x;
        while (its[x]) ++x;
      }
      its[x] = true;
      if (isLucky(n - m + i) && isLucky(n - m + x)) ++ans;
    }
    printf("%d\n", ans);
  }
  return 0;
}
