Problem: You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.

Divisor of n is any such natural number, that n can be divided by it without remainder.

Input

The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).

Output

If n has less than k divisors, output -1.

Otherwise, output the k-th smallest divisor of n.

Examples

Input

4 2


Output

2


Input

5 3


Output

-1


Input

12 5


Output

6

Note

In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.

In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
inline void chmin(A &a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A &a, B b) {
  if (a < b) a = b;
}
const long long M = 31622777;
bool f[M];
vector<pair<long long, long long> > p;
vector<long long> v;
void dfs(long long n, long long m) {
  if (n == p.size()) {
    v.push_back(m);
    return;
  }
  for (long long i = 0; i <= p[n].second; i++) {
    dfs(n + 1, m);
    m *= p[n].first;
  }
}
signed main() {
  long long N, K;
  cin >> N >> K;
  K--;
  fill_n(f, M, 1);
  f[0] = f[1] = 0;
  for (long long i = 2; i < M; i++) {
    if (!f[i]) continue;
    if (N % i == 0) {
      long long cnt = 0;
      while (N % i == 0) {
        N /= i;
        cnt++;
      }
      p.push_back({i, cnt});
    }
    for (long long j = i + i; j < M; j += i) f[j] = 0;
  }
  if (N != 1) p.emplace_back(N, 1);
  dfs(0, 1);
  sort((v).begin(), (v).end());
  if (v.size() > K)
    cout << v[K] << endl;
  else
    cout << -1 << endl;
  return 0;
}