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 PYTHON):

Solution:
import math

def solution(n, k):
  sq = int(math.sqrt(n))

  divs_num = 0
  for i in xrange(1, sq + 1):
    if n % i == 0:
      divs_num += 1

      if i != n / i:
        divs_num += 1

  if divs_num < k:
    print(-1)
    return

  curr = 0
  for i in xrange(1, sq + 1):
    if n % i == 0:
      curr += 1

      if curr == k:
        print(i)
        return
      elif divs_num - curr + 1 == k:
        print(n / i)
        return


n, k = map(int, raw_input().split())
solution(n, k)

# solution(4, 2)
# solution(5, 3)
# solution(12, 5)