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
import math
#import random
#sys.setrecursionlimit(100000000)
input = sys.stdin.readline
 
############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
    return(int(input()))
def inara():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############

n,k=invr()

div=[]

limit=int(math.sqrt(n+1))

for i in range(1,limit+1):
	if n%i==0:
		div.append(i)
		if i*i!=n:
			div.append(n//i)

div.sort()

if len(div)<k:
	print(-1)
else:
	print(div[k-1])