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:
[n, k] = map(int,raw_input().strip().split(' '))

forw = []
back = []

for i in range(1,int(n ** (0.5)) + 2):
  if n % i == 0 and i <= n/i:
    p = n/i
    forw.append(i)
    if p != i:
      back.append(p)

l = len(forw) + len(back)
if l < k:
  print "-1"
else:
  if k <= len(forw):
    print forw[k-1]
  else:
    now = len(forw)
    ans = 0
    ind = -1
    while now != k:
      ans = back[ind]
      ind -= 1
      now += 1
    print ans