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 PYTHON):

Solution:
from math import sqrt
n, k = map(int, raw_input().split())

n_sqrt = sqrt(n)

nf = 0
t = 0
b = []

for i in range(1, int(n_sqrt)+1):


	if n%i == 0:
		t += 1
		b.append(n/i)
		if i*i == n:
			nf += 1
		else:
			nf += 2

		if t == k:
			print i
			exit(0)




# print b, nf
if k > nf:
	print -1
else:
	print b[(nf-k)]