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>
long long int nd[1000000] = {0};
int compare(const void *a, const void *b) {
  if (*(long long int *)a - *(long long int *)b < 0) return -1;
  if (*(long long int *)a - *(long long int *)b > 0) return 1;
  return 0;
}
int main() {
  long long int num, k;
  scanf("%lld %lld", &num, &k);
  int x = sqrt(num);
  int c = 0, i = 0, flag = 1;
  for (i = 1; i <= x; i++) {
    if (num % i == 0) {
      nd[c++] = i;
      if ((long long int)i * i == num) break;
      nd[c++] = num / i;
    }
  }
  qsort(nd, c, sizeof(long long int), compare);
  if (k > c)
    printf("-1");
  else
    printf("%lld", nd[k - 1]);
  return 0;
}