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 sys
from math import*
input=sys.stdin.buffer.readline
import os

n,k=map(int,input().split())

div=[]
count=0
ans=0
last=0
for i in range(1,int(sqrt(n))+1):
	if n%i==0:
		last=i
		count+=1
		if count==k:
			ans=i
			break
		else:
			div.append(n//i)
if ans!=0:
	print(ans)
else:
	if last==div[-1]:
		div.pop()
	div=div[::-1]
#	print(div)
#	print(count)
	if (k-count)<=len(div):
		print(div[k-count-1])
	else:
		print("-1")