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:
def getDivs(n):
    res = [1]
    if n == 1:
        return res
    p = 2

    while p * p <= n:
        if n % p == 0:
            c = p
            res2 = res[:]
            while n % p == 0:
                res.extend([c * x for x in res2])
                c *= p
                n //= p
        p += 1
    if n > 1:
        res.extend([n * x for x in res])
    
    res.sort()
    return res

n, k = map(int, input().split())
ds = getDivs(n)

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