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 PYTHON3):

Solution:
arr = [int(x) for x in input().split()]
n = arr[0]
k = arr[1]
div = [] 
p = int(pow(n, 0.5)) + 1
for x in range(1, p):
  if (n % x == 0):
    div.append(x)
    if (n//x != x):
      div.append(n//x)
div = list(sorted(div))
if k > len(div):
  print (-1)
else:
  print (div[k-1])