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:
from math import sqrt,ceil,floor

arr = [int(i) for i in input().split()]
n = arr[0]
k = arr[1]

count = 0

arr = []

for i in range(1,floor(sqrt(n))+1):
	if (n%i == 0):
		count += 1
		arr.append(i)


found = 0
# print(arr)
arrs = arr[::-1]

if (count == 0):
	print(-1)
else:
	if (len(arr)>=k):
		print (arr[k-1])
		found = 1
	else:
		for i in arrs:
			if (n//i == i):
				continue
			else:
				arr.append(n//i)
				# print(arr)
				if (len(arr)==k):
					print (arr[k-1])
					found = 1
					break

if (found == 0):
	print(-1)