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 sys
def get_ints(): return map(int, sys.stdin.readline().strip().split())

#def get_list(): return list(map(int, sys.stdin.readline().strip().split()))

import math
#for ad in range(int(input())):
n,k=get_ints()
ans=-1;c=1
if n==1 and k==1:ans=1

for i in range(2,int(n**0.5)+1):
    if n%i==0:
        c+=1
    if c==k:
        ans=i
        break
if k==1:ans=1
if ans!=-1 or n==1:
    print(ans)
else:
    xx=0
    if int(n**0.5)**2==n:
        xx=1

    for i in range( int(n ** 0.5)-xx ,0,-1):
        if n % i == 0:
            c += 1
        if c == k:
            ans = n//i
            break
    print(ans)