# Codeforces869a the artful expedient

Posted by yjjr's blog on February 6, 2018

Rock... Paper!

After Karen have found the deterministicwinning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comesup with a new game as a substitute. The game works as follows.

A positive integer n is decided first.Both Koyomi and Karen independently choose n distinct positive integers,denoted byx1, x2, ..., xn and y1, y2, ..., yn respectively. They reveal their sequences, and repeat until all of 2nintegers become distinct, which is the only final state to be kept andconsidered.

Then they count the number of ordered pairs(i, j) (1 ≤ i, j ≤ n) such that the valuexixor yj equals to one of the 2n integers. Here xormeans thebitwiseexclusive or operation on two integers, and is denoted by operators ^and/or xor in most programming languages.

Karen claims a win if the number of suchpairs is even, and Koyomi does otherwise. And you're here to help determine thewinner of their latest game.

Input

The first line of input contains a positiveintegern (1 ≤ n ≤ 2 000) — the length of both sequences.

The second line containsnspace-separated integers x1, x2, ..., xn (1 ≤ xi ≤ 2·106) —the integers finally chosen by Koyomi.

The third line containsnspace-separated integers y1, y2, ..., yn (1 ≤ yi ≤ 2·106) —the integers finally chosen by Karen.

Input guarantees that the given 2nintegers are pairwise distinct, that is, no pair (i, j) (1 ≤ i, j ≤ n) exists such that one of the following holds: xi = yj; i ≠ j andxi = xj; i ≠ j and yi = yj.

Output

Output one line — the name of the winner,that is, "Koyomi" or "Karen" (without quotes). Please beaware of the capitalization.

Examples

Input

3
1 2 3
4 5 6

Output

Karen

Input

5
2 4 6 8 10
9 7 5 3 1

Output

Karen

Note

In the first example, there are 6 pairssatisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an evennumber.

In the second example, there are 16 suchpairs, and Karen wins again.

Code

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define LL long long
using namespace std;
const int maxn=2006,maxm=2000006;
int a[maxn],b[maxn];
bool num[maxm];
int n,cnt=0;
int main()
{
scanf("%d",&n);
rep(i,1,n)scanf("%d",&a[i]),num[a[i]]=true;
rep(i,1,n)scanf("%d",&b[i]),num[b[i]]=true;
rep(i,1,n)
rep(j,1,n){
if((a[i]^b[j])>=maxm)continue;
if(num[a[i]^b[j]])cnt++;
}
if(cnt%2==0)cout<<"Karen\n";
else cout<<"Koyomi\n";
return 0;
}