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 
p=input().split()
m=int(p[0])
p=int(p[1]) 
def kthdivisor(n,k) : 
    list = []  
    for i in range(1, int(math.sqrt(n) + 1)) : 
          
        if (n % i == 0) : 
            if (n / i == i) : 
                list=list+[i]
            else :  
                list=list+[i]
                list=list+[int(n/i)]
    l=sorted(list)
    if(len(l)<k):
        print(-1)
    else:
        print(l[k-1])
kthdivisor(m,p)