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:
from math import ceil
n, k = map(int, input().split())
arr = []
fl = 0
if  int(n ** 0.5) ** 2 == n:
    fl = 1
for i in range(1, int(n ** 0.5) + 1):
    if n % i == 0:
        arr.append(i)
if n == k == 1:
    print(1)
elif k > len(arr) * 2 - fl:
    print(-1)
else:
    if k <= len(arr):
        print(arr[k - 1])
    else:
        print(n // arr[2 * len(arr) - k - fl])