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;
const long long inf = 1000000001;
const long long maxg = 4444444444LL;
vector<long long> g;
void recurs(int d, int len, long long z) {
  if (d == len)
    g.push_back(z);
  else {
    recurs(d + 1, len, z * 10 + 4);
    recurs(d + 1, len, z * 10 + 7);
  }
}
long long a[1000];
long long f[100];
long long fact(int n) {
  if (n > 50) return inf;
  if (f[n] != 0) return f[n];
  if (n == 1 || n == 0) return f[n] = 1;
  return f[n] = min(inf, n * min(inf, (fact(n - 1))));
}
bool good(int a) {
  if (a == 0) return false;
  while (a > 0) {
    if (!(a % 10 == 4 || a % 10 == 7)) return false;
    a /= 10;
  }
  return true;
}
int main() {
  for (int ll = 1; ll <= 9; ll++) {
    recurs(0, ll, 0);
  }
  g.push_back(maxg);
  long long n, k;
  cin >> n >> k;
  if (fact(n) < k) {
    cout << -1 << endl;
    return 0;
  }
  long long p = 1;
  long long m = n - 1;
  while (fact(n - m) + 1 <= k) m--;
  long long s = n - m;
  for (int i = 1; i <= s; i++) a[i] = m + i;
  while (p < k) {
    for (int i = 1;; i++) {
      if (fact(i + 1) + p > k) {
        p += fact(i);
        for (int j = s - i + 1;; j++) {
          if (a[j] > a[s - i]) {
            swap(a[j], a[s - i]);
            break;
          }
        }
        break;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < int(g.size()); i++) {
    if (g[i] <= m) ans = i + 1;
  }
  for (int i = 1; i <= s; i++) {
    if (good(i + m) && good(a[i])) ans++;
  }
  cout << ans << endl;
  return 0;
}
