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;
long long fac[14];
vector<long long> lucky;
bool isLucky(long long n) {
  int x;
  while (n) {
    x = n % 10;
    n /= 10;
    if (x == 4 || x == 7) continue;
    return 0;
  }
  return 1;
}
void gen(long long n) {
  long long x = n * 10;
  if (x > 1000000000ll) return;
  x += 4;
  lucky.push_back(x);
  gen(x);
  x += 3;
  lucky.push_back(x);
  gen(x);
}
int main() {
  int i, j, r, ans;
  long long n, k, m;
  cin >> n >> k;
  fac[0] = 1;
  for (i = 1; i < 14; i++) fac[i] = i * fac[i - 1];
  if (n < 14 && fac[n] < k) {
    cout << -1;
    return 0;
  }
  gen(0);
  lucky.push_back(4444444444ll);
  sort(lucky.begin(), lucky.end());
  m = max(n - 13, 0ll);
  ans = upper_bound(lucky.begin(), lucky.end(), m) - lucky.begin();
  vector<bool> used(n - m + 1, false);
  for (i = n - m; i >= 1; i--) {
    r = (k + fac[i - 1] - 1) / fac[i - 1];
    k -= (r - 1) * fac[i - 1];
    for (j = 1;; j++)
      if (!used[j]) {
        r--;
        if (!r) break;
      }
    used[j] = 1;
    if (isLucky(n - i + 1) && isLucky(m + j)) ans++;
  }
  cout << ans;
  return 0;
}
