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
from math import ceil
if __name__ == "__main__" :
    n , k = map(int, input().split())  
    L = ceil(sqrt(n)) + 1
    totalDivisors = 0 
    for i in range(1, L) :
        if n % i == 0 :
             
            if (n//i == i) :
                totalDivisors += 1
                break
            elif (n/i < i) :
                break
            else :
                totalDivisors += 2
            
    if k > totalDivisors :
        print(-1) 
    else :
        j = totalDivisors + 1
        m = 0
        for i in range(1, L) :
            if (n % i == 0) :
                m += 1
                j -= 1
                #print("M AND J", m , j , "I , N//I"  , i , n//i )

                if (m == k) :
                    print(i)
                    break  
                elif(j == k) :
                    print(n//i)
                    break  
                else :
                    pass