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 collections
import itertools

n, k = map(int, input().split())
def prime_factors(n):
    i = 2
    while i * i <= n:
        if n % i == 0:
            n //= i
            yield i
        else:
            i += 1

    if n > 1:
        yield n


def prod(iterable):
    result = 1
    for i in iterable:
        result *= i
    return result


def get_divisors(n):
    pf = prime_factors(n)

    pf_with_multiplicity = collections.Counter(pf)

    powers = [
        [factor ** i for i in range(count + 1)]
        for factor, count in pf_with_multiplicity.items()
    ]

    for prime_power_combo in itertools.product(*powers):
        yield prod(prime_power_combo)

res = list(get_divisors(n))
res.sort()
if len(res) >= k:
    print(res[k - 1])
else:
    print(-1)