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 ceil,sqrt
def divisors(n,k):
    begin= []
    end = []
    for i in range(1, int(sqrt(n))+1):
        if n % i is 0:
            if n//i is i:
                begin.append(i)
            else:
                begin.append(i)
                end.append(n//i)
    begin += end[::-1]
    size = len(begin)
    if size < k:
        print(-1)
    else:
        print(begin[k-1])

n,k = input().split()
n,k = int(n), int(k)
divisors(n,k)