Codeforces Round #439 (Div. 2) A题 he Artful Expedien

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 choosen distinct positive integers, denoted by x₁, x₂, ..., x_n andy₁, y₂, ..., y_n respectively. They reveal their sequences, and repeat untilall 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 valuex_ixor y_j equals to one of the2n 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 integersx₁, x₂, ..., x_n (1 ≤ x_i ≤ 2·10⁶) — the integers finally chosen by Koyomi.

The third line contains n space-separated integersy₁, 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

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.

题目意思就是 给你两组数据 让a[i]^b[j] 如果这个数在a并b这个集合中出现过的话，那么计数器+1 最后判断判断奇偶对应输出即可， 做题时候看着数据不大，就2000 个,n^2暴力跑一下，这样写法虽然能ac 但是本着优化的原则，在赛后经过大家的一番讨论，惊奇地发现，这道题目只输出"Karen"即可。比赛时候 我看到有人这么做了，我还想去hack人家的， 结果 呵呵了， 还是自己太菜。  终于这样为什么能ac呢？  假设 x属于a，y属于b，如果存在 z 属于a并b，则根据题意，x^y=z。 既然 这个成立， 那么 x^z = y也将成立，这样的话，对于每一组满足题意的数据， 那么一定存在另外一组与之对应的也满足题意。

这里 给出 x^y=z则x^z=y的证明。 方程两同时xor x，则 x^y^x=z^x。 化简右边，得 y=x^z。 即证。

代码很简单， 这里就不贴了。  好题哇！