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 P[(100)], g[(100)];
vector<long long> f;
int len;
long long n, m;
bool vis[(100)];
void dfs(long long x, int t) {
  if (t > 10) return;
  f.push_back(x);
  dfs(x * 10ll + 4ll, t + 1);
  dfs(x * 10ll + 7ll, t + 1);
}
bool ok(long long x) {
  while (x > 0ll) {
    if (x % 10ll != 4ll && x % 10ll != 7ll) return false;
    x /= 10ll;
  }
  return true;
}
long long find(long long pos) {
  if (n - pos < (long long)len) {
    int x = len - (int)(n - pos);
    return (long long)g[x] + n - (long long)len;
  }
  return pos;
}
int main() {
  dfs(0ll, 0);
  sort(f.begin(), f.end());
  P[0] = 1ll;
  for (long long i = 1ll; i <= 15ll; ++i) P[i] = P[i - 1] * i;
  cin >> n >> m;
  len = 1;
  while (m > P[len]) len++;
  long long tmp = m;
  for (int i = 1; i <= len; ++i) {
    int x = (int)((tmp - 1ll) / P[len - i]), cnt = 0;
    for (int j = 1; j <= len; ++j)
      if (!vis[j]) {
        if (cnt++ >= x) {
          vis[j] = true, g[i] = j;
          break;
        }
      }
    tmp -= (long long)x * P[len - i];
  }
  long long ans = 0ll;
  for (int i = 1; i < f.size(); ++i) {
    if (f[i] > n) break;
    if (ok(find(f[i]))) ans++;
  }
  cout << ans;
  return 0;
}
