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

def calculate(n, k):
    ctr = 0
    a= set()
    p = int(math.sqrt(n))
    for i in range(1,p+1, 1):
        if n % i == 0:
            a.add(i)
            a.add(int(n/i))
    b = list(a)
    b.sort();
##    print(b)
    if len(b) >=  k:
        return b[k-1]
    else:
        return -1

##    for j in range(p,1,-1):
##        if n % j == 0:
##            r = n/j
##            ctr += 1
##            if ctr ==k:
##                return r
##    return -1

a = input().split()
c = calculate(int(a[0]), int(a[1]))
print(c)
##
##print(calculate(4,2))
##print(calculate(5,3))
##print(calculate(12,5))
##