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;
int getall(long long x, int n) {
  if (x > n) return 0;
  return 1 + getall(x * 10 + 4, n) + getall(x * 10 + 7, n);
}
bool is_good(int x) {
  if (!x) return 0;
  while (x) {
    if (x % 10 != 4 && x % 10 != 7) return 0;
    x /= 10;
  }
  return 1;
}
vector<int> kth(int n, int k) {
  vector<long long> f(n);
  f[0] = 1;
  for (auto i = 1; i < n; i++) f[i] = i * f[i - 1];
  vector<int> perm(n), u(n, 0);
  for (auto i = 0; i < n; i++) {
    for (auto j = 0; j < n; j++) {
      if (u[j]) continue;
      long long add = f[n - 1 - i];
      if (add < k) {
        k -= add;
        continue;
      }
      perm[i] = j;
      break;
    }
    u[perm[i]] = 1;
  }
  return perm;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  long long fac = 1;
  int cnt;
  for (auto i = 1; i < n + 1; i++) {
    fac *= i;
    cnt = i;
    if (fac >= k) break;
  }
  if (fac < k) {
    cout << -1;
    return 0;
  }
  int tot = getall(0, n) - 1;
  vector<int> perm = kth(cnt, k);
  for (auto i = 0; i < cnt; i++) {
    int pos = n - cnt + 1 + i, val = perm[i] + n - cnt + 1;
    if (is_good(val) && !is_good(pos)) --tot;
  }
  cout << tot;
}
