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 N_MAX = 16;
const int DEPTH_MAX = 11;
long long FACTORIAL[N_MAX];
vector<long long> luckies;
long long N, K;
bool is_lucky(long long x) {
  if (x == 4 || x == 7) return true;
  if (x % 10 != 4 && x % 10 != 7) return false;
  return is_lucky(x / 10);
}
void make_lucky(long long num, int depth) {
  luckies.push_back(num);
  if (depth == DEPTH_MAX) return;
  make_lucky(10 * num + 4, depth + 1);
  make_lucky(10 * num + 7, depth + 1);
}
vector<int> get_kth(int n, long long k) {
  k--;
  vector<int> remaining;
  vector<int> perm;
  for (int i = 0; i < n; i++) remaining.push_back(i);
  for (int pos = 0; pos < n; pos++) {
    int ix = k / FACTORIAL[n - pos - 1];
    k %= FACTORIAL[n - pos - 1];
    perm.push_back(remaining[ix]);
    remaining.erase(remaining.begin() + ix);
  }
  return perm;
}
int main() {
  make_lucky(0LL, 0);
  FACTORIAL[0] = 1LL;
  for (int i = 1; i < N_MAX; i++) FACTORIAL[i] = i * FACTORIAL[i - 1];
  cin >> N >> K;
  if (N < N_MAX && K > FACTORIAL[N]) {
    cout << "-1\n";
    return 0;
  }
  int lucky_count = 0;
  int perm_len = (int)min(N, (long long)N_MAX);
  vector<int> perm = get_kth(perm_len, K);
  for (int i = 0; i < perm_len; i++) {
    if (is_lucky(N - perm_len + i + 1) && is_lucky(N - perm_len + perm[i] + 1))
      lucky_count++;
  }
  if (N > N_MAX) {
    for (int i = 1; i < luckies.size(); i++) {
      if (luckies[i] <= N - N_MAX) lucky_count++;
    }
  }
  cout << lucky_count << '\n';
}
