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 = map(int, input().split())
on = n
e = 0
while n & 1 == 0:
    n >>= 1
    e += 1
div = []
for i in range(1, int(sqrt(n))+1, 2):
    if n%i == 0:
        for j in range(e+1):
            div.append(i*pow(2, j))
            if n//i != i:
                div.append(n//i*pow(2, j))

if k > len(div):
    print(-1)
elif k == len(div):
    print(on)
elif k == 1:
    print(1)
else:
    div.sort()
    print(div[k-1])