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:
def div(x):
  i=1;ans=[]
  while i*i<=n:
   if x%i==0 and i!=x//i:ans.append(i);ans.append(x//i)
   elif x%i==0:ans.append(i)
   i+=1
  return ans
n,k=map(int,input().split())
ans=div(n)
if len(ans)<k:print(-1)
else:ans=sorted(ans);print(ans[k-1])