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:
from math import sqrt
n,k = [int(i) for i in input().split()]
l=[]
hashl=[]
for i in range(int(sqrt(n))+2):
    hashl.append(0)
for i in range(1,int(sqrt(n))+2):
    if (n%i==0):
        if (hashl[i]==0):
            if (i!=(n//i)):
                l.append(i)
                l.append(n//i)
            else:
                l.append(i)                
        hashl[i]=1
        if (n//i<int(sqrt(n))+2):
            hashl[n//i]=1
l.sort()
length=len(l)
if (length<k):
    print (-1)
else:
    print (l[k-1])