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 n, k;
long long fact[20];
inline long long XF(int x) { return x > 13 ? 10000000000LL : fact[x]; }
int L, R;
map<int, int> pv;
vector<int> avb;
int lucky(int w) {
  while (w) {
    int T = w % 10;
    if (T != 4 && T != 7) return 0;
    w /= 10;
  }
  return 1;
}
int main() {
  cin >> n >> k;
  fact[0] = 1;
  for (int i = 1; i <= 13; ++i) fact[i] = i * fact[i - 1];
  if (XF(n) < k) {
    cout << "-1\n";
    return 0;
  }
  R = n;
  L = R - min(n, 13LL) + 1;
  for (int i = L; i <= R; ++i) avb.push_back(i);
  for (int i = L; i <= R; ++i) {
    int j = 0;
    for (j = 0;; ++j)
      if (XF(R - i) < k)
        k -= XF(R - i);
      else
        break;
    pv[i] = avb[j];
    avb.erase(avb.begin() + j);
  }
  int ans = 0;
  for (int len = 1; len <= 9; ++len)
    for (int msk = 0; msk < (1 << len); ++msk) {
      long long T = 0;
      for (int i = 0; i < len; ++i) T = T * 10 + ((msk & (1 << i)) ? 4 : 7);
      if (T <= n) {
        if (!pv.count(T))
          ans += lucky(T);
        else
          ans += lucky(pv[T]);
      }
    }
  cout << ans << '\n';
  return 0;
}
