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())
l=int(sqrt(n))
lis=[]
ind=0
for i in range(1, l+1):
	if (n%i==0):
		if (i==n//i):
			lis.insert(ind, i)
			ind+=1
			continue
		lis.insert(ind, i)
		lis.insert(ind+1, n//i)
		ind+=1
temp=len(lis)
if (k>temp):
	print(-1)
else:
	print(lis[k-1])