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 bisect
from bisect import bisect_left as lb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
sflo=lambda:list(map(float,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**6)
mod=10**9+7
mod1=998244353
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs(root,par):
    if par!=-1:
        dp[root]=dp[par]+1
    for i in range(1,20):
        if kthpar[root][i-1]!=-1:
            kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
    for child in gp[root]:
        if child==par:continue
        kthpar[child][0]=root
        dfs(child,root)
ans=0
def hnbhai(tc):
    n,k=sd()
    abe=[]
    for i in range(1,int(n**0.5)+1):
        if n%i==0:
            abe.append(n//i)
            abe.append(i)
    abe=list(set(abe))
    abe.sort()
    if len(abe)<k:
        print(-1)
        return
    print(abe[k-1])
for _ in range(1):
    hnbhai(_+1)