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;
long long int n, k, i;
int main(void) {
  scanf("%I64d %I64d", &n, &k);
  float lim = sqrt(n);
  std::vector<long long int> v;
  for (i = 1; i <= lim; ++i) {
    if (n % i == 0) v.push_back(i);
  }
  long long int q = v.size();
  if (n == 1) {
    switch (k) {
      case 1:
        cout << 1;
        break;
      default:
        cout << -1;
        break;
    }
    return 0;
  }
  if (q == 1) {
    switch (k) {
      case 1:
        printf("1");
        break;
      case 2:
        printf("%I64d", n);
        break;
      default:
        printf("-1");
        break;
    }
    return 0;
  }
  if (v[q - 1] * v[q - 1] == n) {
    if (k <= q)
      printf("%I64d", v[k - 1]);
    else if (k < (2 * q) && v[q - 1 - k % q])
      printf("%I64d", n / v[q - 1 - k % q]);
    else
      printf("-1");
  } else {
    if (k <= q)
      printf("%I64d", v[k - 1]);
    else if (k < (2 * q) && v[q - k % q])
      printf("%I64d", n / v[q - k % q]);
    else if (k == 2 * q)
      printf("%I64d", n);
    else
      printf("-1");
  }
  return 0;
}