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;
vector<int> lucky;
void init() {
  lucky.push_back(0);
  for (size_t i = 0; i < lucky.size(); i++) {
    if (lucky[i] >= 100000000) break;
    lucky.push_back(lucky[i] * 10 + 4);
    lucky.push_back(lucky[i] * 10 + 7);
  }
  lucky.erase(lucky.begin());
}
bool islucky(int x) {
  do
    if (x % 10 != 7 && x % 10 != 4) return false;
  while (x /= 10);
  return true;
}
int main() {
  init();
  int n, m, len = 0, ans = 0;
  long long s = 1;
  scanf("%d%d", &n, &m);
  while (s < m) s *= ++len;
  if (n < len) return puts("-1") & 0;
  vector<int> u;
  for (int i = 1; i <= len; i++) u.push_back(n - len + i);
  for (int i = 1; i <= len; i++) {
    s /= len + 1 - i;
    int x = (m - 1) / s;
    if (islucky(n - len + i) && islucky(u[x])) ans++;
    u.erase(u.begin() + x);
    m -= x * s;
  }
  ans += upper_bound(lucky.begin(), lucky.end(), n - len) - lucky.begin();
  printf("%d\n", ans);
}
