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 const maxn = 15;
int a[maxn];
int good[1500];
int sz = 0;
void add_good(int mask, int len) {
  int cur = 0;
  for (int i = len - 1; i >= 0; --i)
    cur = cur * 10 + ((mask & (1 << i)) ? 7 : 4);
  good[sz++] = cur;
}
void generate_good() {
  for (int i = 1; i <= 9; ++i) {
    int l = 1 << i;
    for (int mask = 0; mask < l; ++mask) add_good(mask, i);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  generate_good();
  int n, k;
  cin >> n >> k;
  long long curk = 1;
  int curn = 1;
  while (curk < k) {
    curk *= ++curn;
  }
  --k;
  for (int i = 0; i < maxn - curn; ++i) a[i] = i;
  if (curn > n) {
    cout << -1 << '\n';
    return 0;
  };
  for (; curn > 0;) {
    curk /= curn--;
    int need = k / curk;
    k %= curk;
    int curnumber = -1;
    while (need >= 0) {
      ++curnumber;
      bool flag = 1;
      for (int i = 0; i < maxn; ++i)
        if (a[i] == curnumber) flag = 0;
      need -= flag;
    }
    a[maxn - curn - 1] = curnumber;
  }
  int ans = 0;
  for (int i = 0; i < sz; ++i) {
    if (good[i] <= n) {
      int from = good[i];
      int to = from;
      if (n - to < maxn) to = a[maxn - 1 - (n - to)] + n - maxn + 1;
      if (binary_search(good, good + sz, to)) ++ans;
    }
  }
  cout << ans << endl;
  return 0;
}
