Codeforces Round #439 (Div. 2)

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, ..., xnand 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 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

31 2 34 5 6

output

Karen

input

52 4 6 8 109 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.

题意：

从上排和下排各找一对，他们的异或还可以在这两排找到。若对数为偶：karen。

POINT：

可以暴力做，数组要开成2·106 *2，因为有些数异或出来会更大。

也可以直接输出karen，因为a^b=c,c属于两排之间，那么肯定存在a^c=b或者b^c=a。

那么答案肯定为偶数。

注意数中没有0，所以不用多虑。不可能有异或出来相等的情况。

#include <iostream>#include <string.h>#include <stdio.h>#include <math.h>#include<algorithm>using namespace std;#define LL long longconst int maxn = 2222;int x[maxn];int y[maxn];int flag[4000000+44];int main(){    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++){        scanf("%d",&x[i]);        flag[x[i]]=1;    }    for(int i=1;i<=n;i++){        scanf("%d",&y[i]);        flag[y[i]]=1;    }    int ans=0;    for(int i=1;i<=n;i++){        for(int j=1;j<=n;j++){            if(flag[x[i]^y[j]]==1){                ans++;            }        }    }    if((ans%2)==0){        printf("Karen\n");    }else{        printf("Koyomi\n");    }}