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 isLucky(long long n) {
  if (n == 0) return false;
  while (n) {
    if (n % 10 != 4 && n % 10 != 7) return false;
    n /= 10;
  }
  return true;
}
long long fact[20];
long long cnt(long long k, vector<long long>& n, long long curInd) {
  if (n.size() == 0) return 0;
  long long sz = n.size();
  long long cur = 0;
  while (fact[sz - 1] <= k) {
    k -= fact[sz - 1];
    cur++;
  }
  long long res = 0;
  if (isLucky(n[cur]) && isLucky(curInd)) res = 1;
  n.erase(n.begin() + cur);
  curInd++;
  return res + cnt(k, n, curInd);
}
int main() {
  fact[0] = fact[1] = 1;
  for (long long i = 2; i < 20; ++i) {
    fact[i] = fact[i - 1] * i;
  }
  long long n, k;
  cin >> n >> k;
  if (n < 20 && k > fact[n]) {
    cout << -1 << endl;
    return 0;
  }
  k--;
  long long res = 0;
  for (long long len = 1;; ++len) {
    for (long long i = 0; i < (1 << len); ++i) {
      long long num = 0;
      for (long long j = len - 1; j >= 0; --j) {
        num *= 10;
        num += ((i >> j) & 1) ? 7 : 4;
      }
      if (num > (n - 15)) goto next;
      res++;
    }
  }
next:;
  vector<long long> lastN;
  long long start;
  for (long long i = n; i >= 1 && i >= (n - 15); --i) {
    lastN.push_back(i);
  }
  start = lastN.back();
  reverse(lastN.begin(), lastN.end());
  cout << res + cnt(k, lastN, start) << endl;
  return 0;
}
