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 math
d = []
def divisors(n, k):
    p = 0
    for i in range(1, int(math.sqrt(n))+1):
        if n%i == 0:
            d.append(i)
            if n//i==i:
                p = 1
            k -= 1
        if k==0:
            return (True, i, p)
    return (False, -1, p)
n, k = map(int, input().split())
(found, ans, parity) = divisors(n, k)
if found:
    print(ans)
else:
    l = len(d)
    if (parity == 0 and k<=2*l):
        print(n//d[2*l-k])
    elif (parity == 1 and k<=2*l-1):
        print(n//d[2*l-1-k])
    else:
        print("-1")