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> lucky;
long long n, k, x, ans;
long long fact[30];
long long ahmad[30];
bool vis[30];
bool ok;
map<long long, long long> lu;
void f(long long x, int i) {
  if (i == 10) return;
  lucky.push_back(x * 10 + 4);
  f(x * 10 + 4, i + 1);
  lu[x * 10 + 4] = 1;
  lucky.push_back(x * 10 + 7);
  f(x * 10 + 7, i + 1);
  lu[x * 10 + 7] = 1;
}
void fa() {
  fact[0] = 1;
  for (long long i = 1; i <= 14; i++) fact[i] = i * fact[i - 1];
}
void get(long long s) {
  for (long long i = 1; i <= s; i++) {
    for (long long j = 1; j <= s; j++) {
      if (k <= fact[s - i] && !vis[j]) {
        vis[j] = 1;
        ahmad[i] = j;
        break;
      } else if (!vis[j])
        k -= fact[s - i];
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  fa();
  if (n < 15 && fact[n] < k) {
    cout << -1;
    return 0;
  }
  f(0, 0);
  sort(lucky.begin(), lucky.end());
  if (n < 14) {
    get(n);
    for (int i = 1; i <= 13; i++) ans += (lu[ahmad[i] + x] && lu[i + x]);
    cout << ans;
    return 0;
  }
  x = n - 12;
  get(13);
  for (long long i = 0; i < lucky.size() && lucky[i] < x; i++) ans++;
  for (int i = 1; i <= 13; i++) ans += (lu[ahmad[i] + x - 1] && lu[i + x - 1]);
  cout << ans;
  return 0;
}
