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;
inline bool isgood(int a) {
  while (a)
    if ((a % 10 != 4) && (a % 10 != 7))
      return false;
    else
      a /= 10;
  return true;
}
inline int countgood(int r) {
  int ans = 0, a = 0, b[10];
  for (int i = 0; i < 10; i++) b[i] = 0;
  for (;;) {
    int i = 0;
    while (b[i] == 7) {
      b[i] = 4;
      i++;
    }
    b[i] = (b[i] == 0) ? 4 : 7;
    for (i = 9, a = 0; i >= 0; i--) a = a * 10 + b[i];
    if (a > r) break;
    ans++;
  }
  return ans;
}
int f[13], v[13], t[13];
int main() {
  f[0] = 1;
  for (int i = 1; i <= 12; i++) f[i] = f[i - 1] * i;
  int n, k;
  cin >> n >> k;
  k--;
  if (n <= 12) {
    if (f[n] <= k) {
      cout << -1;
      return 0;
    }
    for (int i = 0; i < n; i++) v[i] = i + 1;
    for (int i = n; i; i--) {
      int j = k / f[i - 1];
      k %= f[i - 1];
      t[n - i + 1] = v[j];
      for (; j < i - 1; j++) v[j] = v[j + 1];
    }
    int ans = 0;
    for (int i = 1; i <= n; i++)
      if (isgood(i) && isgood(t[i])) ans++;
    cout << ans;
    return 0;
  }
  for (int i = 0; i < 12; i++) v[i] = n - 11 + i;
  for (int i = 12; i; i--) {
    int j = k / f[i - 1];
    k %= f[i - 1];
    t[12 - i + 1] = v[j];
    for (; j < i - 1; j++) v[j] = v[j + 1];
  }
  int ans = 0;
  for (int i = 1; i <= 12; i++)
    if (isgood(n - 12 + i) && isgood(t[i])) ans++;
  ans += countgood(n - 12);
  cout << ans;
  return 0;
}
