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;
vector<long long> lucky;
set<long long> st;
void rec(string s) {
  if (s.size() > 9) return;
  if (s != "") lucky.push_back(atoi(s.c_str()));
  rec(s + "4");
  rec(s + "7");
}
long long n, k;
long long dp[16];
void init() {
  dp[0] = 1;
  for (int i = 1; i < 16; ++i) dp[i] = i * dp[i - 1];
  rec("");
  sort(lucky.begin(), lucky.end());
  st = set<long long>(lucky.begin(), lucky.end());
}
int main() {
  init();
  cin >> n >> k;
  if (n <= 15) {
    if (dp[n] < k) {
      cout << -1 << endl;
      return 0;
    }
  }
  k--;
  vector<int> v;
  for (int i = max(0LL, n - 15); i < n; ++i) v.push_back(i + 1);
  int i = 0;
  while (k > 0) {
    int j;
    for (j = i; k > 0 and j < v.size(); ++j) {
      if ((j - i) * dp[v.size() - i - 1] > k) break;
    }
    --j;
    swap(v[i], v[j]);
    sort(v.begin() + i + 1, v.end());
    k -= (j - i) * dp[v.size() - i - 1];
    ++i;
  }
  int t = n - v.size();
  int ret = 0;
  for (int i = 0; i < lucky.size(); ++i) {
    if (lucky[i] >= t) break;
    ++ret;
  }
  for (int i = 0; i < v.size(); ++i) {
    if (st.count(v[i]) and st.count(t + i + 1)) ++ret;
  }
  cout << ret << endl;
}
