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:16777216")
using namespace std;
vector<long long> v;
long long p[22];
int who[22], what[22];
void dfs(long long n) {
  if (n > 4444444444LL) return;
  v.push_back(n);
  dfs(n * 10 + 4);
  dfs(n * 10 + 7);
}
int main() {
  dfs(4);
  dfs(7);
  sort((v).begin(), (v).end());
  long long n, k, cnt = 1, cnt1 = 1;
  cin >> n >> k;
  for (int i = 2; i <= n; i++) {
    cnt1 = cnt1 * i;
    if (cnt1 >= k) break;
  }
  if (cnt1 < k) {
    cout << -1 << endl;
    return 0;
  }
  p[0] = 1;
  for (int i = 1; i < 15; i++) p[i] = p[i - 1] * i;
  int m;
  for (m = 2;; m++) {
    cnt = cnt * m;
    if (cnt >= k) break;
  }
  if (k == 1) m = 0;
  size_t r = upper_bound((v).begin(), (v).end(), n - m) - v.begin();
  int ans = r;
  long long cnt2 = 0;
  memset(who, -1, sizeof(who));
  memset(what, -1, sizeof(what));
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < m; j++) {
      if (what[j] != -1) continue;
      if (cnt2 + p[m - i - 1] < k)
        cnt2 = cnt2 + p[m - i - 1];
      else {
        who[i] = j;
        what[j] = 0;
        break;
      }
    }
  }
  for (int i = 0; i < m; i++)
    if (binary_search((v).begin(), (v).end(), n - m + i + 1) &&
        binary_search((v).begin(), (v).end(), who[i] + n - m + 1))
      ans++;
  cout << ans << endl;
  return 0;
}
