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 F[20];
vector<int> lucky;
long long n;
void get(long long num = 0, int tam = 0) {
  if (num > n) return;
  if (tam) lucky.push_back(num);
  get(num * 10 + 4, tam + 1);
  get(num * 10 + 7, tam + 1);
}
void permuta(vector<int>& x, long long k) {
  int tam = x.size();
  for (int i = 0, quedan = tam - 1; i < tam; i++, quedan--) {
    sort(x.begin() + i, x.end());
    int mueren = 0;
    while (k > F[quedan]) mueren++, k -= F[quedan];
    swap(x[i], x[i + mueren]);
  }
}
int main() {
  long long k;
  cin >> n >> k;
  vector<int> lasts;
  for (int i = n; lasts.size() < 15 && i >= 1; i--) lasts.push_back(i);
  F[0] = 1;
  for (int i = 1; i < 17; i++) F[i] = F[i - 1] * i;
  if (F[lasts.size()] < k) {
    cout << -1 << endl;
    return 0;
  }
  get();
  sort(lucky.begin(), lucky.end());
  permuta(lasts, k);
  int antes = n - lasts.size();
  int res = 0;
  for (int i = 0; i < lucky.size(); i++)
    if (lucky[i] <= antes)
      res++;
    else {
      int pos = find(lasts.begin(), lasts.end(), lucky[i]) - lasts.begin();
      pos += antes + 1;
      res += binary_search(lucky.begin(), lucky.end(), pos);
    }
  cout << res << endl;
}
