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>
using namespace std;
long long good(long long n) {
  while (n > 0) {
    long long x = n % 10;
    n /= 10;
    if (x == 4 || x == 7) {
    } else
      return 0;
  }
  return 1;
}
signed main() {
  long long n, k;
  cin >> n >> k;
  if (n > 13) {
    cout << "-1\n";
    return 0;
  }
  long long a[n + 1];
  memset(a, 0, sizeof(a));
  long long fact[n + 5];
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) fact[i] = i * fact[i - 1];
  long long sum = 0;
  set<long long> s;
  for (long long i = 1; i <= n; i++) s.insert(i);
  for (long long i = 1; i <= n; i++) {
    long long done = 0;
    for (auto j : s) {
      if (sum + (done + 1) * fact[n - i] < k) {
        done++;
      } else {
        sum += done * fact[n - i];
        s.erase(j);
        a[i] = j;
        break;
      }
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    if (good(i) && good(a[i])) ans++;
  }
  cout << ans;
  return 0;
}
