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 n, k;
long long ln[10000], fac[15];
int cnt, ans, suf;
int tem[15];
void input() { scanf("%d%d", &n, &k); }
void dfs(long long x) {
  if (x > 1000000000) return;
  ln[cnt++] = x;
  dfs(x * 10 + 4);
  dfs(x * 10 + 7);
}
void make_lucky_number() {
  cnt = 0;
  dfs(4);
  dfs(7);
  sort(ln, ln + cnt);
}
void get_my_suf(int s) {
  bool used[15];
  memset(used, false, sizeof(used));
  for (int i = 1; i <= s; ++i) {
    int u = (k - 1) / fac[s - i];
    k -= (u * fac[s - i]);
    u++;
    for (int j = 1; j <= s; ++j) {
      if (!used[j]) u--;
      if (u == 0) {
        used[tem[i] = j] = true;
        break;
      }
    }
  }
}
bool check_lucky(int x) {
  while (x) {
    if (x % 10 != 4 && x % 10 != 7) return false;
    x /= 10;
  }
  return true;
}
void solve() {
  fac[0] = fac[1] = 1;
  for (int i = 2; i <= 13; ++i) fac[i] = fac[i - 1] * i;
  if (n <= 13 && k > fac[n]) {
    puts("-1");
    return;
  }
  for (int i = 0; i < cnt; ++i)
    if (n - 13 < ln[i]) {
      ans = i;
      break;
    }
  suf = min(n, 13);
  get_my_suf(suf);
  for (int i = 1; i <= suf; ++i)
    if (check_lucky(n - suf + tem[i]) && check_lucky(n - suf + i)) ans++;
  printf("%d\n", ans);
}
int main() {
  input();
  make_lucky_number();
  solve();
  return 0;
}
