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 = 1000ll * 1000 * 1000;
const int size = 100;
long long smb, n, k;
int ans;
bool used[size];
int perm[size];
void genthemall(long long num) {
  if (num > inf) return;
  if (num <= n - 15 && num >= 1) ans++;
  genthemall(num * 10 + 4);
  genthemall(num * 10 + 7);
}
bool islucky(long long num) {
  if (num == 0) return false;
  while (num > 0) {
    if (num % 10 != 4 && num % 10 != 7) return false;
    num /= 10;
  }
  return true;
}
int main() {
  cin >> n >> k;
  long long fct = 1;
  int h = min(n, 15ll), i, j;
  for (i = 1; i <= h; i++) fct *= i;
  if (fct < k) {
    cout << -1 << endl;
    return 0;
  }
  smb = 15;
  ans = 0;
  genthemall(0);
  for (i = 0; i < h; i++) used[i] = false;
  for (i = 0; i < h; i++) {
    fct /= (h - i);
    for (j = 0; j < h; j++)
      if (!used[j])
        if (fct >= k) {
          used[j] = true;
          perm[i] = j;
          break;
        } else
          k -= fct;
    ans += islucky(n - h + i + 1) && islucky(n - h + perm[i] + 1);
  }
  cout << ans << endl;
  return 0;
}
