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;
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const double EPS = 1e-9;
const double PI = acos(-1.);
const long long LL_INF = 1e17 + 16;
const int INF = 1e9 + 10;
const long long MOD = 1e9 + 7;
const int MAXN = 20;
set<long long> goods;
long long f[MAXN];
void gen() {
  for (int i = (1); i <= (int)(10); i++) {
    for (int j = (0); j < (int)(1 << i); j++) {
      long long v = 0;
      long long st10 = 1;
      for (int k = (0); k < (int)(i); k++) {
        if (j & (1 << k)) {
          v += 7 * st10;
        } else {
          v += 4 * st10;
        }
        st10 *= 10;
      }
      goods.insert(v);
    }
  }
}
int per[MAXN];
int used[MAXN];
void solve() {
  gen();
  long long n, k;
  cin >> n >> k;
  f[0] = 1;
  int m;
  for (m = 1;; m++) {
    f[m] = f[m - 1] * m;
    if (f[m] >= k) {
      break;
    }
  }
  if (m > n) {
    printf("-1\n");
    return;
  }
  int ans = 0;
  int offset = n - m;
  for (__typeof((goods).begin()) it = (goods).begin(); it != (goods).end();
       it++) {
    if ((*it) <= offset) {
      ans++;
    }
  }
  k--;
  int top = 1;
  for (int i = m; i >= 1; i--) {
    int v = k / f[i - 1] + 1;
    int cnt = 0;
    for (int j = (1); j <= (int)(m); j++) {
      if (!used[j]) {
        cnt++;
        if (cnt == v) {
          per[top++] = j;
          used[j] = 1;
          break;
        }
      }
    }
    k %= f[i - 1];
  }
  for (int i = (1); i <= (int)(m); i++) {
    if (goods.count(i + offset) && goods.count(per[i] + offset)) {
      ans++;
    }
  }
  cout << ans << endl;
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
}
