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 int N = 2048;
bool use[20];
long long a[N];
long long ff(long long k) {
  long long res = 0;
  while (k > 1) {
    if (k % 2 == 0)
      res = res * 10 + 4;
    else
      res = res * 10 + 7;
    k /= 2;
  }
  return res;
}
int check(int d) {
  for (int i = 0; i < N; i++)
    if (a[i] == d) return 1;
  return 0;
}
int main() {
  long long n, k, f = 1, i, x, j, f1, d, t, kol, i1;
  int res = 0;
  cin >> n >> k;
  for (i = 0; i < N; i++) {
    a[i] = ff(i + 1);
  }
  sort(a, a + N);
  if (n <= 14) {
    for (i = 2; i <= n; i++) f *= i;
    if (k > f) {
      cout << -1;
      return 0;
    }
  }
  f = 1;
  i = 1;
  while (f < k) f *= ++i;
  x = n - i;
  t = i;
  j = 1;
  while (a[j] <= x) {
    j++;
    res++;
  }
  while (i > 0) {
    f1 = f / i;
    d = (k + (f1 - 1)) / f1;
    kol = 0;
    for (i1 = 0; i1 < t; i1++) {
      if (!use[i1]) kol++;
      if (kol == d) break;
    }
    use[i1] = true;
    k -= (d - 1) * f1;
    res += (check(i1 + x + 1) & check(n - i + 1));
    i--;
    f = f1;
  }
  cout << res;
  return 0;
}
