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
import math
import bisect

def solve(n, k):
    A = []
    for i in range(1, math.floor(math.sqrt(n)) + 1):
        if n % i == 0:
            A.append(i)
            j = n // i
            if j != i:
                A.append(j)
    if len(A) < k:
        return -1
    A.sort()
    return A[k-1]

def main():
    n, k = map(int, input().split())
    ans = solve(n, k)
    print(ans)

if __name__ == "__main__":
    main()