Problem Description

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, ..., x_n and y₁, y₂, ..., y_n 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 x_i xor y_j equals to one of the 2n integers. Here xor 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, ..., x_n (1 ≤ x_i ≤ 2·10⁶) — the integers finally chosen by Koyomi.

The third line contains n space-separated integers y₁, y₂, ..., y_n (1 ≤ y_i ≤ 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: x_i = y_j; i ≠ j and x_i = x_j; i ≠ j and y_i = y_j.

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.

这题的题意确实挺难看懂……

大意是给a,b两个序列。求对于a,b序列中各自任意取出一个数，并对他们进行异或操作。

将异或完的值与a,b序列中的值进行比较。如果a,b中有与其相同的值，则cnt++。

cnt代表了有几个值相同。

如果cnt为偶数，则Karen赢，如果cnt为奇数，则Koyomi赢。

需要注意的是两个值异或后可能会超过原先定的2*10⁶这个上限许多。（例如1111111111^0000000001 ?）

所以vis数组需要开的大点。（2倍？）

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
int n,vis[5000005],a[2005],b[2005],cnt;
char num[25]; 
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        vis[a[i]]=1;
    }
    for(int i=0;i<n;i++){
        scanf("%d",&b[i]);
        vis[b[i]]=1;
    }
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            if(vis[a[i]^b[j]])
                cnt++;
    if(cnt&1)
        printf("Koyomi");
    else
        printf("Karen");
}