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:
from math import sqrt

def main():
    (n, k) = map(int, input().split(' '))
    if k > n:
        return -1
    s = set()
    a = []
    b = []
    for i in range(1, int(sqrt(n)) + 1):
        if n % i == 0:
            if i not in s:
                a.append(i)
                s.add(i)
            if n // i not in s:
                b.append(n // i)
                s.add(n // i)
    a = a + b[::-1]
    ret = -1
    try:
        ret = a[k - 1]
    except Exception as e:
        ret = -1
    return ret

print(main())