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;
bool islucky(int x) {
  if (x % 10 != 4 || x % 10 != 7) return false;
  if (x < 10) return true;
  return islucky(x / 10);
}
int main() {
  int N, k;
  scanf("%d%d", &N, &k);
  vector<int> u;
  for (int(i) = 0; (i) < (int)(10); ++(i))
    for (int(mask) = 0; (mask) < (int)(1 << i); ++(mask)) {
      int p = 0;
      for (int(j) = 0; (j) < (int)(i); ++(j))
        p = p * 10 + (mask >> j & 1) ? 7 : 4;
      u.push_back(p);
    }
  int n = min(N, 13);
  vector<long long> f(20, 1);
  for (int(i) = (int)(1); (i) <= (int)(19); ++(i)) f[i] = f[i - 1] * 1ll * i;
  if (k > f[n]) return puts("-1");
  vector<bool> vis(n, false);
  vector<int> v;
  for (int(i) = 0; (i) < (int)(n); ++(i))
    for (int(j) = 0; (j) < (int)(n); ++(j))
      if (!vis[j]) {
        if (k < f[n - 1 - i]) {
          v.push_back(j);
          vis[j] = true;
          break;
        } else
          k -= f[n - 1 - i];
      }
  int ans = 0, m = N - n + 1;
  for (int(i) = 0; (i) < (int)(n); ++(i))
    if (islucky(m + i) && islucky(m + v[i])) ans++;
  for (int(i) = 0; (i) < (int)(u.size()); ++(i))
    if (u[i] < m) ans++;
  printf("%d\n", ans);
}
