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 = map(int, input().split())

sqrt = int(math.sqrt(n) + 1)

is_square = sqrt * sqrt == n

divisors = []
for i in range(1, sqrt):
    if n % i == 0:
        divisors.append(i)
        if i * i != n:
            divisors.append(n // i)

divisors.sort()

if k > len(divisors):
    print(-1)
else:
    print(divisors[k - 1])