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


def printDivisors(n):
    l = []
    i = 1
    while (i * i < n):
        if (n % i == 0):
            l.append(i)
            if (len(l) >= k):
                break
        i += 1

    for i in range(int(n ** 0.5), 0, -1):
        if (n % i == 0):
            l.append(n // i)
            if (len(l) >= k):
                break
    return l


res = printDivisors(n)
if len(res) >= k:
    print(res[k - 1])
else:
    print(-1)