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:
#####################################
import atexit, io, sys, collections, math, heapq, fractions
buffer = io.BytesIO() 
sys.stdout = buffer
@atexit.register 
def write(): sys.__stdout__.write(buffer.getvalue())
#####################################

def divisors(n):
	r = []
	t = []
	i = 1
	while(i *i <= n):
		if n % i == 0:
			r.append(i)
			if i * i != n: 
				t.append(n/i)
		i+=1
	return r + t[::-1]
n,k = map(int,raw_input().split())
r = divisors(n)
print r[k-1] if k-1 < len(r) else -1