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;
vector<long long> lucky;
void rec(long long x) {
  if (x > 10000000000ll) return;
  lucky.push_back(x);
  rec(x * 10ll + 4);
  rec(x * 10ll + 7);
}
int main() {
  long long f[14];
  f[0] = 1;
  for (int i = 1; i < 14; i++) f[i] = f[i - 1] * (long long)i;
  int n, k;
  cin >> n >> k;
  if (n <= 13 && k > f[n]) {
    cout << -1 << endl;
    return 0;
  }
  int ans = 0;
  rec(4), rec(7);
  sort(lucky.begin(), lucky.end());
  for (int i = 0; i < (int)lucky.size(); i++)
    if (lucky[i] <= n - 13) ans++;
  --k;
  int mask = (1 << 13) - 1;
  string last;
  for (int i = 0; i < min(n, 13); i++) {
    int get = min(n, 13) - i - 1, j;
    for (j = 0; j <= get && k >= f[get]; j++) k -= f[get];
    for (int k = 0; k < 13; k++) {
      if (mask & (1 << k)) {
        if (!j) {
          mask ^= (1 << k);
          last += k + '1';
          break;
        } else
          j--;
      }
    }
  }
  for (int i = 0; i < min(n, 13); i++)
    if (count(lucky.begin(), lucky.end(), n - i) &&
        (last[(int)last.size() - i - 1] == '4' ||
         last[(int)last.size() - i - 1] == '7'))
      ans++;
  cout << ans << endl;
  return 0;
}
