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 n, k;
vector<int> v(0);
long long f[14];
vector<long long> lucky(0);
int main() {
  f[0] = 1;
  for (int i = 1; i <= 13; ++i) f[i] = f[i - 1] * (long long)i;
  cin >> n >> k;
  if (n < 14) {
    if (k > f[n]) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int len = 1; len <= 10; ++len)
    for (int mask = 0; mask < (1 << len); ++mask) {
      long long val = 0;
      for (int(i) = 0; (i) < (len); ++(i))
        if (mask & (1 << i))
          val = val * 10 + 4;
        else
          val = val * 10 + 7;
      lucky.push_back(val);
    }
  int start = max(n - 12, 1);
  int ans = 0;
  for (int(i) = 0; (i) < ((int)(lucky).size()); ++(i))
    if (lucky[i] < start) ans++;
  v.clear();
  for (int val = start; val <= n; ++val) v.push_back(val);
  int perm[13];
  int len = n - start;
  int iter = 0;
  for (int step = start; step < n; ++step) {
    int pos = 0;
    while (k > f[len]) {
      k -= f[len];
      pos++;
    }
    perm[iter++] = v[pos];
    v.erase(v.begin() + pos);
    len--;
  }
  perm[iter++] = v[0];
  iter = 0;
  for (int step = start; step <= n; ++step) {
    if (find((lucky).begin(), (lucky).end(), step) != lucky.end())
      if (find((lucky).begin(), (lucky).end(), perm[iter]) != lucky.end())
        ans++;
    iter++;
  }
  cout << ans << endl;
}
