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:
def main():
    def div(n):
        if n==1:
            return [1]
        if n==2:
            return [1,2]
        if n==3:
            return [1,3]
        K=int(n**0.5)
        temp1=[]
        temp2=[]
        for i in range(1,K+1):
            if n%i==0:
                temp1+=[i]
                temp2+=[n//i]
        if temp1[-1]==temp2[-1]:
            temp2=temp2[:-1]
        temp=temp1+temp2[::-1]
        return temp
    n,k=list(map(int,input().strip().split(' ')))
    D=div(n)
    if len(D)<k:
        print(-1)
    else:
        print(D[k-1])
main()