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>
#pragma comment(linker, "/STACK:134217728")
using namespace std;
vector<long long> lucky;
long long fact[21];
void next_add(long long q) {
  if (q > 10000000000ll) return;
  lucky.push_back(q * 10ll + 4ll);
  lucky.push_back(q * 10ll + 7ll);
  next_add(q * 10ll + 4ll);
  next_add(q * 10ll + 7ll);
}
vector<int> kth_perm(int n, long long k) {
  if (n == 1) return vector<int>(1, 0);
  long long first = k / fact[n - 1];
  vector<int> ans(1, first);
  vector<int> add = kth_perm(n - 1, k % fact[n - 1]);
  for (int i(0); i <= (((int)add.size()) - 1); ++i) {
    if (add[i] >= first) ++add[i];
    ans.push_back(add[i]);
  }
  return ans;
}
bool is_lucky(int n) {
  if (find((lucky).begin(), (lucky).end(), n) != lucky.end()) return true;
  return false;
}
int main() {
  fact[0] = fact[1] = 1ll;
  for (int i(0); i <= (19 - 1); ++i)
    fact[i + 2] = fact[i + 1] * (long long)(i + 2);
  int pos, n, k;
  scanf("%d%d", &n, &k);
  if (n < 21 && fact[n] < (long long)k) {
    printf("-1\n");
    return 0;
  }
  for (pos = 0; fact[pos] <= k; ++pos)
    ;
  vector<int> d;
  for (int i = n - pos + 1; i <= n; ++i) d.push_back(i);
  int res(0);
  next_add(0);
  sort((lucky).begin(), (lucky).end());
  res +=
      lower_bound((lucky).begin(), (lucky).end(), n - pos + 1) - lucky.begin();
  vector<int> kth = kth_perm(pos, (long long)(k - 1));
  vector<int> perm(pos);
  for (int i(0); i <= (pos - 1); ++i) perm[i] = d[kth[i]];
  for (int i(0); i <= (pos - 1); ++i)
    if (is_lucky(perm[i]) && is_lucky(n - pos + 1 + i)) ++res;
  printf("%d\n", res);
  return 0;
}
