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;
long long factorial(long long n) {
  long long p = 1;
  for (int i = 1; i <= n; ++i) {
    p *= i;
  }
  return p;
}
int main() {
  long long n, k;
  cin >> n >> k;
  if (n <= 12) {
    int s = 1;
    for (int i = 1; i <= n; ++i) {
      s *= i;
    }
    if (k > s) {
      cout << -1;
      return 0;
    }
  }
  set<int> happy;
  for (int i = 1; i <= 9; ++i) {
    vector<int> a(i + 1, 0);
    while (a[0] == 0) {
      int h = 0;
      for (int j = i; j > 0; --j) {
        h *= 10;
        h += (a[j] == 0 ? 4 : 7);
      }
      happy.insert(h);
      for (int j = i;; --j) {
        if (a[j] == 1) {
          a[j] = 0;
        } else {
          a[j] = 1;
          break;
        }
      }
    }
  }
  if (k == 1) {
    int count = 0;
    set<int>::iterator it;
    for (it = happy.begin(); it != happy.end(); ++it) {
      if (*it <= n) {
        ++count;
      }
    }
    cout << count;
    return 0;
  }
  int i = n;
  while (factorial(n - i) < k) {
    --i;
  }
  int count = 0;
  set<int>::iterator it;
  for (it = happy.begin(); it != happy.end(); ++it) {
    if (*it <= i) {
      ++count;
    }
  }
  set<int> numbers;
  for (int j = i + 1; j <= n; ++j) {
    numbers.insert(j);
  }
  for (int j = i + 1; j <= n; ++j) {
    int x = (k - 1) / factorial(n - j);
    set<int>::iterator it = numbers.begin();
    for (int l = 0; l < x; ++l) {
      ++it;
    }
    if ((happy.find(j) != happy.end()) && (happy.find(*it) != happy.end())) {
      ++count;
    }
    numbers.erase(*it);
    k -= x * factorial(n - j);
  }
  cout << count;
  return 0;
}
