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;
string lower(string s) {
  for (int i = 0; i < s.size(); i++) s[i] = tolower(s[i]);
  return s;
}
vector<string> sep(string s, char c) {
  string temp;
  vector<string> res;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == c) {
      if (temp != "") res.push_back(temp);
      temp = "";
      continue;
    }
    temp = temp + s[i];
  }
  if (temp != "") res.push_back(temp);
  return res;
}
template <class T>
T toint(string s) {
  stringstream ss(s);
  T ret;
  ss >> ret;
  return ret;
}
template <class T>
string tostr(T inp) {
  string ret;
  stringstream ss;
  ss << inp;
  ss >> ret;
  return ret;
}
template <class T>
void D(T A[], int n) {
  for (int i = 0; i < n; i++) cout << A[i] << " ";
  cout << endl;
}
template <class T>
void D(vector<T> A, int n = -1) {
  if (n == -1) n = A.size();
  for (int i = 0; i < n; i++) cout << A[i] << " ";
  cout << endl;
}
long long fact[21];
set<long long> lnum;
int main() {
  for (int(i) = (1); (i) <= (10); (i)++) {
    int mask = (1 << i);
    for (int(j) = (0); (j) < (mask); (j)++) {
      long long tnum = 0;
      for (int(k) = (0); (k) < (i); (k)++) {
        tnum *= 10LL;
        if ((j & (1 << k)))
          tnum += 4LL;
        else
          tnum += 7LL;
      }
      lnum.insert(tnum);
    }
  }
  fact[0] = fact[1] = 1;
  for (int(i) = (2); (i) < (21); (i)++) fact[i] = (long long)i * fact[i - 1];
  int n, k;
  cin >> n >> k;
  int ans = 0;
  for (typeof(lnum.begin()) it = lnum.begin(); it != lnum.end(); it++) {
    if (*it > n) break;
    if (n - *it + 1 <= 15) break;
    ans++;
  }
  int sn = max(1, n - 15 + 1);
  if (fact[n - sn + 1] >= k) {
    int totaliter = n - sn + 1;
    set<int> poss;
    for (int(i) = (sn); (i) <= (n); (i)++) poss.insert(i);
    for (int(i) = (0); (i) < (totaliter); (i)++) {
      set<int>::iterator it = poss.begin();
      while (k > fact[totaliter - i - 1]) {
        it++;
        k -= fact[totaliter - i - 1];
      }
      if (lnum.find(*it) != lnum.end() && lnum.find(sn + i) != lnum.end())
        ans++;
      poss.erase(it);
    }
    cout << ans << endl;
  } else {
    cout << -1 << endl;
  }
  return 0;
}
