Codeforces Round #439 (Div. 2) The Artful Expedient

A. The Artful Expedient

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Rock... Paper!

After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up 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 2n integers become distinct,
which is the only final state to be kept and considered.

Then they count the number of ordered pairs (i, j) (1 ≤ i, j ≤ n)
such that the value xi xor yj equals
to one of the 2n integers. Herexor means
the bitwise exclusive 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 such pairs is even, and Koyomi does otherwise. And you're here to help determine the winner of their latest game.

Input

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

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

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

Input guarantees that the given 2n integers 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 and xi = xj; i ≠ j and yi = yj.

Output

Output one line — the name of the winner, that is, "Koyomi" or "Karen"
(without quotes). Please be aware 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 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3).
Thus, Karen wins since6 is an even number.

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

这道题题意就是给你两串数字，让你记录上面一行的任意数字与下面一行任意数字进行异或操作的结果依然在所给数组中的个数。

根据个数的奇偶性来输出不同的结果。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
using namespace std;
int a[2020];
int b[2020];
bool v[2222222];
int main()
{

int n;
while(cin>>n)
{
memset(v,0,sizeof(v));
for(int i=0;i<n;i++)
{
cin>>a[i];
v[a[i]]=1;
}
for(int i=0;i<n;i++)
{
cin>>b[i];
v[b[i]]=1;
}
int res=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
if(v[a[i]^b[j]])
res++;
}
}
if(res%2==1)
cout<<"Koyomi"<<endl;
else
cout<<"Karen"<<endl;
}
return 0;
}