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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int M = 4e5 + 3;
vector<long long> lucky;
void gen() {
  for (int i = 1; i <= 9; i++) {
    for (int j = 0; j < (1 << i); j++) {
      long long cur = 0;
      for (int k = 0; k < i; k++) {
        if (j & (1 << k))
          cur = cur * 10 + 7;
        else
          cur = cur * 10 + 4;
      }
      lucky.push_back(cur);
    }
  }
  sort(lucky.begin(), lucky.end());
}
bool isLucky(long long x) {
  while (x) {
    if (x % 10 != 4 and x % 10 != 7) return false;
    x /= 10;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  gen();
  long long n, k, c = 0;
  cin >> n >> k;
  vector<long long> f, p;
  k--;
  while (k) {
    if (c + 1 > n) return cout << "-1", 0;
    f.push_back(k % (c + 1));
    p.push_back(n - c);
    k /= c + 1;
    c++;
  }
  reverse(p.begin(), p.end());
  long long ans =
      upper_bound(lucky.begin(), lucky.end(), n - c) - lucky.begin();
  for (int i = c - 1; i >= 0; i--) {
    if (isLucky(n - i) and isLucky(p[f[i]])) ans++;
    p.erase(p.begin() + f[i]);
  }
  cout << ans;
}
