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 double eps = 1e-8;
const long long lloo = 1e18;
const int oo = 1e9;
int islucky(int x) {
  while (x) {
    if (x % 10 != 4 && x % 10 != 7) return 0;
    x /= 10;
  }
  return 1;
}
int solve(int d) {
  int res = 0;
  for (long long m = 1, _b = 9; m <= _b; m++) {
    int l = 1 << m;
    for (long long i = 0, _b = l - 1; i <= _b; i++) {
      int x = i;
      int y = 0;
      for (int j = 0; j < m; j++) {
        if (x & 1)
          y = y * 10 + 7;
        else
          y = y * 10 + 4;
        x >>= 1;
      }
      if (y <= d) res++;
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  int gt[15];
  gt[0] = 1;
  for (long long i = 1, _b = 12; i <= _b; i++) gt[i] = gt[i - 1] * i;
  int d = -1;
  for (long long i = 12, _b = 1; i >= _b; i--)
    if (gt[i] < k) {
      d = i;
      break;
    }
  if (n <= d) {
    cout << -1 << "\n";
    return 0;
  }
  int res = solve(n - d - 1);
  vector<int> v;
  for (long long i = n - d, _b = n; i <= _b; i++) v.push_back(i);
  k--;
  int pos = n - d;
  for (long long i = d, _b = 0; i >= _b; i--) {
    int x = k / gt[i];
    if (islucky(pos) && islucky(v[x])) res++;
    v.erase(v.begin() + x);
    k %= gt[i];
    pos++;
  }
  cout << res << "\n";
}
