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:
inp = list(map(int,input().split()))
n=inp[0]
k=inp[1]
if n==1:
    if k==1:
        print (1)
    else:
        print (-1)
else:
    arr=[1]
    arr1=[n]
    for i in range(2,int(pow(n,0.5))+1):
        if n%i==0:
            arr.append(i)
            if i!=n//i:
                arr1.append(n//i)
    flag=0
    if len(arr)+len(arr1)<k:
        print (-1)
        flag=1
    if k<=len(arr) and flag==0:
        print (arr[k-1])
        flag=1
    elif k>len(arr) and flag==0:
        k-=len(arr)
        print (arr1[len(arr1)-k])