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:
n,k = input().split()
n,k = int(n),int(k)
l1=[1]
if n != 1 :
    l2 = [n]
else :
    l2 = []
i = 2
while i*i <= n :
    if n%i == 0 :
        l1.append(i)
        if i*i != n :
            l2.append(int(n/i))

    i += 1
l2.reverse()
l = l1+l2
if k> len(l) :
    print(-1)    
else :
    print(l[k-1])