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:
#!/usr/bin/env python3

inf = float('inf')
read_ints = lambda : list(map(int, input().split()))

def solve(n, k):
    ls = []
    ncnt = 0
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            ncnt += 1
            ls.append(i)
            if ncnt == k:
                return i
    if ls[-1]**2 == n:
        if k >= 2*ncnt:
            return -1
        return n // ls[2 * ncnt-1 - k]
    else:
        if k > 2*ncnt:
            return -1
        return n // ls[2 * ncnt - k]

if __name__ == '__main__':
    n, k = read_ints()
    print(solve(n, k))