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 PYTHON):

Solution:
from math import sqrt

n, k = map(int, raw_input().split())
q = []
sn = sqrt(n)
for i in range(1, int(sn) + 1):
    if n % i == 0:
        q.append(i)
        if len(q) == k:
            print(i)
            break
else:
    if sn == int(sn):
        if k <= len(q) * 2 - 1:
            print(n // q[-(k-len(q)+1)])
        else:
            print(-1)
    else:
        if k <= len(q) * 2:
            print(n // q[-(k-len(q))])
        else:
            print(-1)