codeforces—— 869A —— 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 by x1, 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. Here xormeans
the bitwise exclusive or operation on two integers, and is denote
cfc8
d 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 since 6 is an even number.

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

上面的数和下一个数两两组合, 每一个组合中 如果异或后的值 在输入中出现过.ans++,ans为偶数Karen赢,反之另一位赢..

事实上,只有Karen会赢,因为 如果a^b=c,则abc中任意两个元素组合,则可以获得另外一个元素..(写一下存在c,在输入中的情况就容易理解了)

没有考虑上面的问题的话,就是普通的暴力,也不慢....15ms.

#include <cstring>
#include <queue>
#include <cstdio>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <iostream>
#include <cmath>
using namespace std;
const int len=1e6;
int IN[4*len+8];
bool com[2*len+4];
int main() {
int n;
while(~scanf("%d",&n)) {
for(int i=0; i<n; i++)
scanf("%d",&IN[i]);
for(int i=0; i<n; i++)
scanf("%d",&IN[n+i]);
memset(com,false,sizeof com);
for(int i=0; i<2*n; i++)
com[IN[i]]=true;
int ans=0;
for(int i=0; i<n; i++)
for(int j=0; j<n; j++) {
int t=IN[i]^IN[n+j];
if(t<=2*len)
ans+=com[t];
}
if(ans&1) printf("Koyomi\n");
else printf("Karen\n");
}
return 0;
}