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;
const long double PI = 3.14159265358979323;
const long long int linf = 1000111000111000111LL;
const long long int inf = 1011111111;
const long long int N = 100005;
vector<long long int> v1[10], v2;
set<long long int> st2;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int test = 1;
  while (test--) {
    long long int n, k;
    cin >> n >> k;
    long long int fac[13];
    fac[0] = 1;
    for (long long int i = 1; i <= 12; i++) fac[i] = fac[i - 1] * i;
    if (n <= 12) {
      if (k > fac[n]) {
        cout << "-1\n";
        continue;
      }
    }
    v1[1].push_back(4);
    v1[1].push_back(7);
    for (long long int i = 2; i <= 9; i++) {
      for (long long int j = 0; j < (long long int)(v1[i - 1]).size(); j++) {
        v1[i].push_back(v1[i - 1][j] * 10 + 4);
        v1[i].push_back(v1[i - 1][j] * 10 + 7);
      }
    }
    for (long long int i = 1; i <= 9; i++) {
      for (long long int j = 0; j < (long long int)(v1[i]).size(); j++) {
        v2.push_back({v1[i][j]});
        st2.insert(v1[i][j]);
      }
    }
    long long int ans = 0;
    if (n > 13) {
      long long int ele = n - 13;
      long long int p1 =
          upper_bound((v2).begin(), (v2).end(), ele) - v2.begin();
      p1--;
      ans += p1;
    }
    set<long long int> st1;
    for (long long int i = n; i >= max((long long int)1, n - 13); i--)
      st1.insert(i);
    long long int pos = max((long long int)1, n - 13);
    while (!st1.empty()) {
      long long int len = (long long int)(st1).size();
      long long int q1 = fac[len - 1];
      long long int num = k / q1;
      if (k % q1 != 0) num++;
      auto it = st1.begin();
      for (long long int i = 0; i < num - 1; i++) it++;
      long long int ele = *it;
      st1.erase(ele);
      if (st2.find(ele) != st2.end() and st2.find(pos) != st2.end()) ans++;
      pos++;
      k = k % q1;
      if (k == 0) k = q1;
    }
    cout << ans << "\n";
  }
  cerr << '\n'
       << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms\n";
}
