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 solve(n, k):
    cnt = 0
    A = []
    for i in range(1, int(n ** 0.5) + 1):
        if n % i == 0:
          cnt += 1
          A.append(i)
        if cnt == k:
            return i
    bol = n // A[-1] == A[-1]
    if ((cnt - bol) * 2 + bol) >= k: return n // A[cnt - bol - k]
    return -1
 
n, k = map(int, input().split())
 
print(solve(n, k))