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;
bool islucky(int n) { return n == 4 || n == 7; }
int main() {
  vector<long long> fs(20, 1);
  for (int i = int(1); i < int(20); i++) fs[i] = fs[i - 1] * i;
  vector<long long> ls;
  for (int i = 1; i <= 10; i++) {
    for (int j = int(0); j < int(1 << i); j++) {
      long long n = 0;
      for (int k = int(0); k < int(i); k++) {
        n *= 10;
        if (1 << k & j)
          n += 7;
        else
          n += 4;
      }
      ls.push_back(n);
    }
  }
  sort((ls).begin(), (ls).end());
  for (int n, k; cin >> n >> k && n | k;) {
    if (n < 20 && fs[n] < k) {
      cout << -1 << endl;
      continue;
    }
    int len = upper_bound((fs).begin(), (fs).end(), k) - fs.begin();
    vector<int> a(len);
    iota((a).begin(), (a).end(), 0);
    for (int i = int(0); i < int(len); i++) {
      int j = upper_bound((fs).begin(), (fs).end(), k) - fs.begin() - 1;
      if (j == -1) break;
      int p = k / fs[j];
      rotate(a.begin() + i, a.begin() + i + p, a.begin() + i + p + 1);
      k -= fs[j] * p;
    }
    int res = 0;
    for (int i = int(0); i < int(ls.size()); i++)
      if (ls[i] <= n - len) res++;
    for (int i = int(0); i < int(len); i++)
      if (binary_search((ls).begin(), (ls).end(), n - len + a[i] + 1) &&
          binary_search((ls).begin(), (ls).end(), n - len + i + 1))
        res++;
    cout << res << endl;
  }
  return 0;
}
