Codeforces 869A. The Artful Expedient

A. The Artful Expedient

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 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, …, 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.

---

题目的意思是给出2n个不同的数字问上下两行两两异或结果有出现的对数是奇还是偶

思路：直接先存下每个数字数否出现 然后暴力

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>#include <bitset>using namespace std;int a[2005], b[2005], cnt[5000006];int main(){    int n;    memset(a, 0, sizeof a);    memset(b, 0, sizeof b);    memset(cnt, 0, sizeof cnt);    scanf("%d", &n);    for (int i = 0; i < n; i++)        scanf("%d", &a[i]), cnt[a[i]]++;    for (int i = 0; i < n; i++)        scanf("%d", &b[i]), cnt[b[i]]++;    int ans = 0;    for (int i = 0; i < n; i++)        for (int j = 0; j < n; j++)            if (cnt[a[i] ^ b[j]] > 0)                ans++;    printf("%s\n", ans % 2 ? "Koyomi" : "Karen");    return 0;}