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=(int(i) for i in input().split())
if(k==1):
    print(1)
else:
    l=[1]
    c=1
    k1=0
    k2=2
    for i in range(2,int(math.sqrt(n))+1):
        if(n%i==0):
            c+=1
            l+=[i]
            k2+=2
        if(c==k):
            print(i)
            k1=1
            break
    if(k1==0):
        if(math.sqrt(n)==int(math.sqrt(n))):
            k2-=1
        if(k2<k):
            print(-1)
        else:
            if(k2%2==0):
                print(n//l[c-k+c])
            else:
                print(n//l[c-k+c-1])