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 n, k;
bool isLucky(long long x) {
  if (!x) return 1;
  if (!(x % 10 == 7 || x % 10 == 4)) return 0;
  return isLucky(x / 10);
}
set<int> st;
int luck(long long x) {
  if (x > n) return 0;
  long long cnt = 0;
  if (x) cnt = (st.find(x) == st.end());
  cnt += luck(x * 10 + 4);
  cnt += luck(x * 10 + 7);
  return cnt;
}
vector<int> get_kth(vector<int> a, int k) {
  int nn = a.size();
  vector<int> idx(nn);
  int div = 1;
  for (int place = 1; place <= nn; place++) {
    if ((k / div) == 0) break;
    idx[n - place] = (k / div) % place;
    div *= place;
  }
  for (int i = 0; i < nn; i++) {
    int id = idx[i] + i;
    if (id != i) {
      int tmp = a[id];
      for (int j = id; j > i; j--) a[j] = a[j - 1];
      a[i] = tmp;
    }
  }
  return a;
}
void exist() {
  long long cur = 1;
  for (long long i = 1; i <= n; i++) {
    cur *= i;
    if (cur >= k) return;
  }
  puts("-1");
  exit(0);
}
int solve(vector<int> a) {
  k--;
  long long offset = max(0ll, n - 15);
  int ret = 0;
  a = get_kth(a, k);
  for (int i = 0; i < a.size(); i++) {
    ret += (isLucky(a[i]) & isLucky(offset + 1 + i));
  }
  return ret;
}
int main() {
  scanf("%lld%lld", &n, &k);
  exist();
  vector<int> v;
  for (int i = n; i > max(0ll, n - 15); i--) v.push_back(i), st.insert(i);
  reverse(v.begin(), v.end());
  cout << luck(0) + solve(v);
  return 0;
}
