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 long long INF = 10000000000LL;
long long first[22];
int ans;
int get(int n, int k) {
  first[0] = 1;
  int i;
  for (i = 1; i <= n && first[i] < k; i++) {
    first[i] = first[i - 1] * i;
    if (first[i] >= k) return i;
  }
  return -1;
}
void luc(long long x, int t) {
  if (x > t) return;
  if (x != 0) ans++;
  luc(x * 10 + 4, t);
  luc(x * 10 + 7, t);
}
bool lucky(int v) {
  while (v) {
    if (v % 10 != 7 && v % 10 != 4) return false;
    v /= 10;
  }
  return true;
}
set<int> second;
int main() {
  int n, k, p, x;
  cin >> n >> k;
  p = get(n, k);
  if (p == -1) {
    cout << -1;
    return 0;
  }
  x = n - p;
  luc(0LL, x);
  for (int i = x + 1; i <= n; i++)
    for (int j = x + 1; j <= n; j++)
      if (second.find(j) == second.end()) {
        if (k <= first[n - i]) {
          second.insert(j);
          if (lucky(i) && lucky(j)) ans++;
          break;
        } else
          k -= first[n - i];
      }
  cout << ans;
  return 0;
}
