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())
i=1
cont=0
flag=1
l=[]
while i*i<=n:
    if n%i==0:
        cont+=1
        l.append(i)
    if cont==k:
        print(i)
        flag=0
        break
    i+=1
if flag:
    z=cont
    for i in range(z-1,-1,-1):
        if n//l[i]!=l[i]:
            l.append(n//l[i])
            cont+=1
        if cont == k:
            print(n//l[i])
            flag = 0
            break
    if flag:
        print(-1)