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 inf = 1000000000;
long long n, k, fac[20];
vector<int> v, lucky;
bool isLucky(long long n) {
  while (n) {
    if (n % 10 != 4 && n % 10 != 7) return false;
    n /= 10;
  }
  return true;
}
void generate(int num, int i) {
  if (i == 10) return;
  if (i) lucky.push_back(num);
  generate(num * 10 + 4, i + 1);
  generate(num * 10 + 7, i + 1);
}
int main() {
  cin >> n >> k;
  int from = max(1ll, n - 13);
  for (int i = from; i <= n; i++) {
    v.push_back(i);
  }
  fac[0] = 1;
  for (int i = 1; i <= 15; i++) fac[i] = i * fac[i - 1];
  int ans = 0;
  for (int i = from; i <= n; i++) {
    for (int j = v.size() - 1; j >= 0; j--) {
      if (fac[v.size() - 1] * j < k) {
        if (isLucky(i) && isLucky(v[j])) ans++;
        k -= fac[v.size() - 1] * j;
        v.erase(v.begin() + j);
        break;
      }
    }
  }
  if (k > 1) {
    puts("-1");
    return 0;
  }
  generate(0, 0);
  sort(lucky.begin(), lucky.end());
  cout << ans + lower_bound(lucky.begin(), lucky.end(), from) - lucky.begin()
       << endl;
  return 0;
}
