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:
import math
from sys import stdin
n,m=[int(x)for x in stdin.readline().split()]

f=0
a=[]
count=0
if m>n:
	print(-1)
elif m==n:
	if m==1:
		print(1)
	else:
		print(-1)
else:				
	x=math.sqrt(n)
	x=int(x)
	for i in range(1,x+1):
		if n%i==0:
			a.append(i)
			if i!= n//i:
			    a.append(n//i)
			    count+=2
			else:

			    count+=1
	# print(a,count)
	# b=set(a)
	# a=list(b)
	# print(a,count)		
	if count<m:
		print(-1)
	else: 
		a.sort()

		print(a[m-1])