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 sys import stdin


def factorize(n):  # o(sqr(n))
    c, ans = 1, []
    while (c * c < n):
        if n % c == 0:
            ans.extend([c, n // c])
        c += 1

    if c * c == n:
        ans.append(c)
    return sorted(ans)


n, k = map(int, stdin.readline().split())
try:
    print(factorize(n)[k - 1])
except:
    print(-1)