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;
constexpr int N = 1e5 + 10;
constexpr int MOD = 1e9 + 7;
long long n, k, fact[N], lim, ans;
vector<int> remain;
int build(int p, long long num, int limit) {
  if (p == limit) return num <= lim;
  return build(p + 1, num * 10 + 4, limit) + build(p + 1, num * 10 + 7, limit);
}
bool is(int x) {
  while (x) {
    if ((x % 10) != 4 && (x % 10) != 7) return false;
    x /= 10;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  lim = max(0ll, n - 13);
  fact[0] = 1;
  for (int i = 1; i <= 14; i++) fact[i] = fact[i - 1] * i;
  if (n <= 13 && fact[n] < k) {
    cout << -1 << '\n';
    return 0;
  }
  for (int i = 1; i < 10; i++) ans += build(0, 0, i);
  for (int i = max(1ll, n - 13); i <= n; i++) remain.push_back(i);
  for (int i = max(1ll, n - 13); i <= n; i++) {
    int p = 0;
    while (fact[n - i] < k) {
      k -= fact[n - i];
      p++;
    }
    if (is(i) && is(remain[p])) ans++;
    remain.erase(remain.begin() + p);
  }
  cout << ans << '\n';
  return 0;
}
