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 = [int(i) for i in raw_input().split()]
k = n[1]
c = 0
l = [0 for i in xrange(int(sqrt(n[0]))+3)]
for j in xrange(1,int(sqrt(n[0]))+1):
	if(n[0]%j==0):
		c+=1
		l[c] = j
num = 0
if(int(sqrt(n[0]))**2 == n[0]):
	num = 2*c - 1
else:
	num = 2*c
#print(num)
if(k <= num):
	if(k<=c):
		print(l[k])
	else:
		print(n[0]//l[num - k + 1])
else:
	print(-1)