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

n, k = [int(a) for a in input().split()]

d = []
q = 0
for i in range(1,int(sqrt(n))+1):
    if n % i == 0:
        d.append(i)
        q += 1
        if n//i != i:
            d.append(n//i)
            q += 1

d = list(sorted(d))

if q < k:
    print(-1)
else:
    print(d[k-1])