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 math
n,k = input().split()
n,k = int(n),int(k)
ans = []; ans = []
for i in range(1,int(math.sqrt(n) + 1)):
    if n % i == 0:
        ans.append(i);
        if n/i != i: ans.append(int(n/i));
mid = int(len(ans)/2)
##print(mid)
if len(ans) % 2 != 0: mid += 1;
k -= 1;
##print(ans)
##print(mid)
if len(ans) <= k:
    print(-1)
elif k < mid:
    print(ans[2*k])
else:
    k %= mid
    ##print(k)
    print(ans[2*(int(len(ans)/2) - k - 1) + 1])