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 getfctrlist(n):
    A=[1,n]
    for i in range(2,int(n**0.5)+1):
        if(n%i==0):
            if(n//i == i):
                A.append(i)
            else:
                A.append(i);A.append(n//i)
    A = set(A)
    A = list(A)
    return A

n,k = map(int,input().split())
F = getfctrlist(n)
F.sort()
if(k>len(F)):
    print("-1")
else:
    print(F[k-1])
# priyanshu kumar