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;
const int MAXD = 1000000000;
vector<long long> lucky;
int n, k, lim, a[14];
long long ans, t[14];
void generate(long long x) {
  if (x) lucky.push_back(x);
  if (x < MAXD) {
    generate(x * 10 + 4);
    generate(x * 10 + 7);
  }
}
inline void permutation(int n, int k) {
  bool select[14];
  memset(select, 0, sizeof select);
  for (int i = 0, p; i < n; ++i) {
    for (p = 1;;) {
      for (; p <= n && select[p]; ++p)
        ;
      if (k <= t[n - i - 1]) {
        a[i] = p;
        select[p] = true;
        break;
      } else {
        k -= t[n - i - 1];
        ++p;
      }
    }
  }
}
inline bool find(int x) {
  for (int i = 0; i < int(lucky.size()); ++i)
    if (lucky[i] == x) return true;
  return false;
}
int main(int argc, char* argv[]) {
  t[0] = 1;
  for (int i = 1; i < 14; ++i) t[i] = t[i - 1] * i;
  cin >> n >> k;
  if (n < 13 && k > t[n]) {
    cout << -1 << endl;
    return 0;
  }
  lim = n > 13 ? n - 13 : 0;
  generate(0);
  sort(lucky.begin(), lucky.end());
  for (int i = 0; i < int(lucky.size()); ++i)
    if (lucky[i] <= lim) ++ans;
  permutation(min(13, n), k);
  for (int i = 0; i < min(n, 13); ++i)
    if (find(a[i] + lim) && find(i + lim + 1)) ++ans;
  cout << ans << endl;
}
