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:
import math
n, k = [int(i) for i in input().split()]
answer_list_front = []
answer_list_back = []
check_upper_lim = int(math.sqrt(n))
for i in range(1, check_upper_lim+1):
    if n%i == 0:
        answer_list_front.append(i)
        answer_list_back.append(n//i)
if i == n/i:
    answer_list_back = answer_list_back[:-1]
len_front = len(answer_list_front)
len_back = len(answer_list_back)
if len_front + len_back < k:
    print(-1)
elif len_front >= k:
    print(answer_list_front[k-1])
else:
    print(answer_list_back[len_back+len_front-k])