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
import math
# sys.stdin = open('input.txt','r')
n,k = map(int, input().split())
divs = [0] * (int(math.sqrt(n)) + 1)
d = 0
def solve(n,k):
	global d
	for i in range(1,int(math.sqrt(n))+1):
		if n % i == 0:
			divs[d] = i
			d+=1
			k-=1
			if k == 0:
				return i
	# print(divs,d,k)
	if k > d:
		return -1
	if n // divs[d-1] != divs[d-1]:
		d-=1
		k-=1
	else:
		d-=1
	if k>d:
		return -1
	else:
		return n//divs[d-k]
if n < k:
	print(-1)
else:
	print(solve(n,k))