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:

i = input().split()
a = int(i[0])
b = int(i[1])
x = 1
cont = []
for x in range (1, (int(a**.5)+1)):
    if a%x==0:
        if x != (a//x):
            cont+=[x]
            cont+=[a//x]
        else:
            cont+=[x]
cont=sorted(cont)
#print(cont)
print(-1 if len(cont)<b else cont[b-1])