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;
vector<long long> v, ext;
long long f[20];
bool vis[20];
bool chck(long long j) {
  while (j) {
    if (j % 10 != 4 && j % 10 != 7) break;
    j /= 10;
  }
  return j == 0;
}
int main() {
  long long i, n, k, st, sum, j, pre, sz, ans;
  f[0] = 1;
  for (i = 1; i <= 13; i++) f[i] = f[i - 1] * i;
  cin >> n >> k;
  if (n <= 13 && k > f[n]) k = f[n];
  for (i = 1; f[i] < k; i++)
    ;
  st = n - i + 1;
  sum = k;
  for (j = 1; j <= i; j++) {
    for (k = st; k <= n; k++) {
      if (!vis[k - st]) {
        if (sum - f[i - j] > 0) {
          sum -= f[i - j];
        } else {
          ext.push_back(k);
          vis[k - st] = 1;
          break;
        }
      }
    }
  }
  queue<long long> q;
  q.push(4);
  q.push(7);
  v.push_back(4);
  v.push_back(7);
  while (!q.empty()) {
    pre = q.front();
    q.pop();
    if (pre * 10 + 4 <= 5000000000LL) {
      q.push(pre * 10 + 4);
      q.push(pre * 10 + 7);
      v.push_back(pre * 10 + 4);
      v.push_back(pre * 10 + 7);
    }
  }
  sz = v.size();
  ans = 0;
  for (i = 0; i < sz && v[i] < st; i++) ans++;
  for (i = st; i <= n; i++) {
    if (chck(i)) {
      if (chck(ext[i - st])) ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
