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> luck;
void go(long long a) {
  if (a) luck.push_back(a);
  if (a > (long long)1e16) return;
  go(a * 10LL + 4LL);
  go(a * 10LL + 7LL);
}
int isluck(long long n) {
  if (n <= 0) return 0;
  while (n) {
    if ((n % 10 != 4) && (n % 10 != 7)) return 0;
    n /= 10;
  }
  return 1;
}
vector<long long> res;
vector<long long> ans;
vector<int> us;
long long fact[20];
void gen(int ind, long long k) {
  if (ind == (int)(ans).size()) return;
  for (int i = 0; i < (int)(res).size(); i++) {
    if (!us[i]) {
      if (k < fact[(int)(ans).size() - ind - 1]) {
        us[i] = 1;
        ans[ind] = res[i];
        return gen(ind + 1, k);
      }
      k -= fact[(int)(ans).size() - ind - 1];
    }
  }
}
int main() {
  fact[0] = 1LL;
  for (int i = 0; i < (19); i++) fact[i + 1] = fact[i] * (i + 1LL);
  go(0);
  sort((luck).begin(), (luck).end());
  long long n, k;
  cin >> n >> k;
  long long shift = n - 15;
  for (int i = n; i >= n - 14; i--) {
    res.push_back(i);
    us.push_back(0);
  }
  if (n < 18) {
    if (k > fact[n]) return 0 * printf("-1\n");
  }
  reverse((res).begin(), (res).end());
  ans.resize((int)(res).size());
  int res = 0;
  for (int i = 0; i < ((int)(luck).size()); i++)
    if (luck[i] <= n - 15) res++;
  gen(0, k - 1);
  for (int i = 0; i < ((int)(ans).size()); i++)
    if (isluck(ans[i]) && isluck(shift + i + 1)) res++;
  cout << res << endl;
  return 0;
}
