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.

Example
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.
題意 兩個 n個不同數的集合 A,B ， 對於每對 < i j > ，A[i]^B[i] 的值如果在兩個集合中任意一個，則cnt++，如果cnt為奇數輸出 Koyomi ，偶數就輸出 Karen 。

分析：資料很小，暴力就可以。
一開始用的 map，超時間，改為陣列把，又陣列越界了，兩個在1e6 範圍內的數字相異或值肯定是要大於1e6的。開大點就好，就是讓1e6的二進位制的數位上全變為1，應該就差不多。

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
#define first fi
#define second se
#define  LL long long
#define fread() freopen("in.txt","r",stdin)
#define fwrite() freopen("out.txt","w",stdout)
#define CLOSE() ios_base::sync_with_stdio(false)

const int MAXN = 2e3+10;
const int MAXM = 1e7;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;

int arr[MAXN];
int brr[MAXN];
bool vis[MAXM]={0} ;
int main(){
    CLOSE();
//  fread();
//  fwrite();
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++) {

        scanf("%d",&arr[i]);
        vis[arr[i]]=1;
    }
    for(int i=1;i<=n;i++) {

        scanf("%d",&brr[i]);
        vis[brr[i]] =1;
    }
    int cnt=0;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++) {
            int c=arr[i]^brr[j];
            if(vis[c]) cnt++;
        }
    }
    if(cnt&1) puts("Koyomi");
    else puts("Karen");
    return 0;
}

B The Eternal Immortality

Even if the world is full of counterfeits, I still regard it as wonderful.

Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.

The phoenix has a rather long lifespan, and reincarnates itself once every a! years. Here a! denotes the factorial of integer a, that is, a! = 1 × 2 × … × a. Specifically, 0! = 1.

Koyomi doesn’t care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of b! years, that is, . Note that when b ≥ a this value is always integer.

As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you’re here to provide Koyomi with this knowledge.

Input
The first and only line of input contains two space-separated integers a and b (0 ≤ a ≤ b ≤ 1018).

Output
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.

Example
Input
2 4
Output
2
Input
0 10
Output
0
Input
107 109
Output
2
Note
In the first example, the last digit of is 2;

In the second example, the last digit of is 0;

In the third example, the last digit of is 2.

分析 ：其實只要當便利過程中又0生成我們就可以停止了，之後都會是0，因為此時已經有因子2和5了。
可以想一下，要判斷一個數n的階乘最後一位是不是有零，只要這個數的階乘結果的分解質因數上有2和5就夠了。（一個數n的階乘含有多少個質因子x ，這個我們可以 用
）
扯的遠了，沒有必要，這個題目，我們只要遍歷就行，同時這裡遍歷肯定不會超時，2這個因子過兩個數肯定有，而5這個因子過5個數又肯定會有。

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
#define first fi
#define second se
#define  LL long long
#define fread() freopen("in.txt","r",stdin)
#define fwrite() freopen("out.txt","w",stdout)
#define CLOSE() ios_base::sync_with_stdio(false)

const int MAXN = 1e5;
const int MAXM = 1e6;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;


int main(){
    CLOSE();
//  fread();
//  fwrite();

    LL a,b;cin>>a>>b;
    if(a==b) {
        puts("1");
        return 0;
    }
    LL ans=(a+1)%10;
    for(LL i = a+2;i<=b;i++) {
        ans=(ans%10*i%10)%10;
        if(ans==0) break;
    }
    printf("%lld\n",ans);
    return 0;
}

C The Intriguing Obsession
— This is not playing but duty as allies of justice, Nii-chan!

— Not allies but justice itself, Onii-chan!

With hands joined, go everywhere at a speed faster than our thoughts! This time, the Fire Sisters — Karen and Tsukihi — is heading for somewhere they’ve never reached — water-surrounded islands!

There are three clusters of islands, conveniently coloured red, blue and purple. The clusters consist of a, b and c distinct islands respectively.

Bridges have been built between some (possibly all or none) of the islands. A bridge bidirectionally connects two different islands and has length 1. For any two islands of the same colour, either they shouldn’t be reached from each other through bridges, or the shortest distance between them is at least 3, apparently in order to prevent oddities from spreading quickly inside a cluster.

The Fire Sisters are ready for the unknown, but they’d also like to test your courage. And you’re here to figure out the number of different ways to build all bridges under the constraints, and give the answer modulo 998 244 353. Two ways are considered different if a pair of islands exist, such that there’s a bridge between them in one of them, but not in the other.

Input
The first and only line of input contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 5 000) — the number of islands in the red, blue and purple clusters, respectively.

Output
Output one line containing an integer — the number of different ways to build bridges, modulo 998 244 353.

Example
Input
1 1 1
Output
8
Input
1 2 2
Output
63
Input
1 3 5
Output
3264
Input
6 2 9
Output
813023575
Note
In the first example, there are 3 bridges that can possibly be built, and no setup of bridges violates the restrictions. Thus the answer is 23 = 8.

In the second example, the upper two structures in the figure below are instances of valid ones, while the lower two are invalid due to the blue and purple clusters, respectively.

哎
按照自己翻譯的做這個題，樣例都過不去，真是（菜）。。 最後Google翻譯了一下才明白。看懂題目，應該就不難了。
題意 ： 有三個集合，每個集合中有如干個點，在滿足每個點不會連相同顏色的點的條件下，我們能夠連邊的情況數目。
分析 ： 任意兩個集合之間連邊的時候是不會影響別的集合，所以我們分開求就行。對於任意兩個集合，每個點不能夠連相同的顏色，就是相當於一個點只能夠和一個點一 一對應，從每個集合中取的個數不同，情況數就不同，用組合數 。

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
#define first fi
#define second se
#define  LL long long
#define fread() freopen("in.txt","r",stdin)
#define fwrite() freopen("out.txt","w",stdout)
#define CLOSE() ios_base::sync_with_stdio(false)

const int MAXN = 5000+10;
const int MAXM = 1e6;
const int mod = 998244353 ;
const int inf = 0x3f3f3f3f;

LL fac[MAXN]; // 階乘
LL C[MAXN+100][MAXN+100]; // 組合數
void init(){
    fac[0]=1;
    for(int i=1;i<=MAXN;i++) fac[i]=fac[i-1]*i%mod;
    for(int i=0;i<=MAXN;i++) C[i][0]=1;
    for(int i=1;i<=MAXN;i++){
        for(int j=1;j<=i;j++)
            C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
    }

//    for(int i=1;i<=10;i++){
//        for(int j=1;j<=i;j++)
//            printf("%d ",C[i][j]);
//        puts("");
//    }
} 
LL in[5];
int main(){
    CLOSE();
//  fread();
//  fwrite();
    init();
    cin>>in[1]>>in[2]>>in[3];
    sort(in+1,in+4);
    LL ans1=0,ans2=0,ans3=0;
    for(int i=0;i<=in[1];i++) 
        ans1=ans1%mod+C[in[1]][i]*C[in[2]][i]%mod*fac[i]%mod;
    for(int i=0;i<=in[1];i++) 
        ans2=ans2%mod+C[in[1]][i]*C[in[3]][i]%mod*fac[i]%mod;
    for(int i=0;i<=in[2];i++) 
        ans3=ans3%mod+C[in[2]][i]*C[in[3]][i]%mod*fac[i]%mod;
    printf("%lld\n",ans1*ans2%mod*ans3%mod);
    return 0;
}