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:
import sys
from math import sqrt, floor
from collections import Counter

def factorize(n):
    limit = floor(sqrt(n))
    factor = Counter()

    p = 2
    while n % p == 0:
        factor[p] += 1
        n //= p

    for p in range(3, limit + 1, 2):
        while n % p == 0:
            factor[p] += 1
            n //= p
    if n > 1:
        factor[n] += 1

    return factor

def make_divisors(n):
    result = [1]

    for p, e in factorize(n).items():
        result = [x * p**i for i in range(e + 1) for x in result]

    return sorted(result)

n, k = map(int, input().split())

divisors = make_divisors(n)

if k > len(divisors):
    ans = -1
else:
    ans = divisors[k - 1]

print(ans)