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 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.

附题目链接：http://codeforces.com/problemset/problem/869/A

先贴上暴力代码：  用vis[]标记异或后的值得时候要注意异或后的范围可能会超出2e6；

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e7 + 10;int vis[maxn];int n,a[2010],b[2010];int main(){    while(~scanf("%d",&n)){        for(int i = 1; i <= n; i ++) scanf("%d",&a[i]);        for(int i = 1; i <= n; i ++) scanf("%d",&b[i]);        memset(vis,0,sizeof(vis));        for(int i = 1; i <= n; i ++){            vis[a[i]] ++; vis[b[i]] ++;        }        int cnt = 0;        for(int i = 1; i <= n; i ++){            for(int j = 1; j <=  n; j ++){                if(vis[a[i]^b[j]]) cnt ++;            }        }        if(cnt % 2) printf("Koyomi\n");        else printf("Karen\n");    }    return 0;}

再说下另一种思路：对于任意一个xi ^yj = z ，假设z属于X，令z = xk，即 xi ^ yj = xk, 由异或的原理可得： xk ^ yj = xi。 假设z属于Y，令z=yk，即xi ^ yj = yk => xi ^ yk = yj；

所以只要xi ^yj的结果在X和Y之间，则一定会有偶数个异或对产生。

#include <cstdio>using namespace std;int main(){    printf("Karen\n");    return 0;}