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:
def list_divisors(n):
    divisors = [1]
    half = []
    if n == 1:
        return divisors
    d = 2
    while d*d <= n:
        if n%d == 0:
            divisors.append(d)
        d += 1
    for x in divisors:
        if x == n**(1/2):
            pass
        else:
            half.append(n//x)
    return sorted(divisors + half)

n,k = map(int,input().split())
ans = list_divisors(n)
if (k-1) not in range(len(ans)):
    print("-1")
else:
    print(ans[k-1])