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;
int p[11];
bool used[15];
long long f[15];
vector<int> g;
bool is_good(long long x) {
  if (x < 4) return false;
  while (x > 0) {
    if (x % 10 != 4 && x % 10 != 7) return false;
    x /= 10;
  }
  return true;
}
int main() {
  int n, ans = 0;
  long long k;
  cin >> n >> k;
  f[1] = f[0] = 1;
  for (long long i = 2; i < 15; ++i) f[i] = i * f[i - 1];
  if (n < 14 && k > f[n]) {
    cout << -1;
    return 0;
  }
  while (true) {
    int j = 0;
    p[j]++;
    while (p[j] > 2) {
      p[j] = 0;
      p[++j]++;
    }
    if (p[9] != 0) break;
    bool good = true;
    int cur = 0, mul = 1, cnt = 0;
    for (int i = 0; i < 9; ++i) {
      if (p[i] != 0) {
        cur += mul * (p[i] == 1 ? 4 : 7);
        if (!good) cnt++;
      } else
        good = false;
      mul *= 10;
    }
    if (cnt == 0 && cur > 0 && cur <= n - 14) ans++;
  }
  int len = (n >= 14 ? 14 : n);
  vector<int> perm(len);
  for (int i = 0; i < len; ++i) {
    int dig = -1;
    for (int j = 0; j < len && dig == -1; ++j)
      if (!used[j]) {
        if (k > f[len - i - 1])
          k -= f[len - i - 1];
        else {
          dig = j;
          used[j] = true;
        }
      }
    perm[i] = max(dig, 0) + (n - len) + 1;
  }
  for (int i = 0; i < perm.size(); ++i)
    if (is_good(perm[i]) && is_good(i + (n - len) + 1)) ans++;
  cout << ans;
  return 0;
}
