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:
#!/usr/bin/env python3

from __future__ import absolute_import
import math
from collections import *
from itertools import imap
def ri():
    return imap(int, raw_input().split())

def find_divisor(num):
    d = set()
    for i in xrange(1, int(math.sqrt(num))+1):
        if num%i == 0:
            if num//i == i:
                d.add(i)
            else:
                d.add(i)
                d.add(num//i)
    return d

n, k = ri()

d = list(find_divisor(n))
d.sort()

if len(d) < k:
    print -1
else:
    print d[k-1]