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 pow, sqrt
n, k = map(int, input().split())
ls = []
limit = int(sqrt(n))
for i in range(1, limit + 1):
    if n % i == 0:
        ls.append(i)
length = len(ls)
# print("Length is ", length)
# print("List is ", ls)
if length == 1:
    if k == 1:
        print(1)
    elif k == 2 and n != 1:
        print(n)
    else:
        print(-1)
    exit()
sx = sqrt(n)
if int(sx) != sx:
    if k > 2 * length:
        print(-1)
    else:
        if k <= length:
            print(ls[k - 1])
        else:
            print(int(n / ls[2 * length - k ]))
else:
    # print(ls)
    if k >= 2 * length:
        print(-1)
    else:
        if k <= length:
            print(ls[k - 1])
        else:
            print(int(n / ls[2 * length - k - 1]))

    # if k > 3:
    #     print(-1)
    # elif k == 1:
    #     print(1)
    # elif k == 2:
    #     print(int(sqrt(n)))
    # else:
    #     print(n)