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 int INF = 1000 * 1000 * 1000;
const long long INF64 = 1000LL * 1000LL * 1000LL * 1000LL * 1000LL * 1000LL;
int n;
long long k;
inline bool is_lucky(int x) {
  while (x > 0) {
    if (x % 10 == 4 || x % 10 == 7)
      x /= 10;
    else
      return false;
  }
  return true;
}
void Load() { cin >> n >> k; }
int mark[20];
int a[20];
int pos[20];
long long fac[20];
vector<long long> g;
string s = ".....................";
void rec(int pos) {
  if (pos > 10) return;
  long long x = 0;
  for (int i = 0; i < pos; i++) x = x * 10 + s[i] - '0';
  g.push_back(x);
  s[pos] = '4';
  rec(pos + 1);
  s[pos] = '7';
  rec(pos + 1);
}
void Solve() {
  fac[0] = 1;
  for (int i = 1; i < 15; i++) {
    fac[i] = fac[i - 1] * i;
  }
  if (n < 14 && fac[n] < k) {
    cout << -1;
    return;
  }
  rec(0);
  sort(g.begin(), g.end());
  int ans = 0;
  int m = max(n - 13, 1);
  int sz = 0;
  for (int i = 1; i < (int)g.size(); i++) {
    if (g[i] >= m) break;
    ans++;
  }
  for (int i = m; i <= n; i++) {
    a[i - m] = i;
    sz++;
  }
  for (int i = 1; i <= sz; i++) {
    for (int j = 0; j < sz; j++)
      if (!mark[j]) {
        if (fac[sz - i] < k)
          k -= fac[sz - i];
        else {
          mark[j] = 1;
          if (is_lucky(a[j]) && is_lucky(m + i - 1)) ans++;
          break;
        }
      }
  }
  cout << ans;
}
int main() {
  Load();
  Solve();
  return 0;
}
