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:
from math import sqrt
n,k=map(int,input().split())
if n==1:
    if k==1:
        print(1)
    else:
        print(-1)
else:
    i=2
    t=sqrt(n)
    l=[1]
    l2=[n]
    k-=1
    while i<t:
        
        if n%i==0:
            l.append(i)
            g=n//i
            l2.append(g)
        i+=1
    if i==t:
        l.append(i)
    a=len(l)
    b=len(l2)


    if k<a:
        print(l[k])
        
    elif k<a+b:
        k=k-a+1
        print(l2[-k])
    else:
        print(-1)