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;
bool check(long long x) {
  if (x == 0) return false;
  while (x) {
    if ((x % 10 != 4) && (x % 10 != 7)) return false;
    x /= 10;
  }
  return true;
}
const int max_f = 20;
const int max_num = (int)1e9;
int p[max_f], fact[max_f];
bool used[max_f];
int main() {
  int n, k, f;
  scanf("%d%d", &n, &k);
  vector<long long> v;
  v.push_back(4), v.push_back(7);
  for (int i = 0; i < ((int)(v).size()); ++i) {
    if (v[i] * 10 + 4 <= max_num) v.push_back(v[i] * 10 + 4);
    if (v[i] * 10 + 7 <= max_num) v.push_back(v[i] * 10 + 7);
  }
  fact[0] = 1;
  for (f = 1; fact[f - 1] * f < k; ++f) fact[f] = fact[f - 1] * f;
  if (f > n) {
    printf("-1\n");
    return 0;
  }
  for (int i = 0; i < f; ++i)
    for (p[i] = 0; p[i] < f; ++p[i])
      if (!used[p[i]]) {
        if (k <= fact[f - i - 1]) {
          used[p[i]] = true;
          break;
        }
        k -= fact[f - i - 1];
      }
  int res = 0;
  for (int i = 0; i < ((int)(v).size()); ++i)
    if ((v[i] <= n - f) ||
        ((v[i] <= n) && check(p[v[i] - (n - f + 1)] + (n - f + 1))))
      res++;
  printf("%d\n", res);
  return 0;
}
