Chess For Three

Alex, Bob and Carl will soon participate in a team chess tournament. Since they are all in the same team, they have decided to practise really hard before the tournament. But it's a bit difficult for them because chess is a game for two players, not three.

So they play with each other according to following rules:

• Alex and Bob play the first game, and Carl is spectating;
• When the game ends, the one who lost the game becomes the spectator in the next game, and the one who was spectating plays against the winner.

Alex, Bob and Carl play in such a way that there are no draws.

Today they have played n games, and for each of these games they remember who was the winner. They decided to make up a log of games describing who won each game. But now they doubt if the information in the log is correct, and they want to know if the situation described in the log they made up was possible (that is, no game is won by someone who is spectating if Alex, Bob and Carl play according to the rules). Help them to check it!
 

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the number of games Alex, Bob and Carl played.

Then n lines follow, describing the game log. i-th line contains one integer ai (1 ≤ ai ≤ 3) which is equal to 1 if Alex won i-th game, to 2 if Bob won i-th game and 3 if Carl won i-th game.

Output

Print YES if the situation described in the log was possible. Otherwise print NO.

Sample 1

Input	Output
3 1 1 2	YES

Sample 2

Inputcopy	Outputcopy
2 1 2	NO

Note

In the first example the possible situation is:

1. Alex wins, Carl starts playing instead of Bob;
2. Alex wins, Bob replaces Carl;
3. Bob wins.

The situation in the second example is impossible because Bob loses the first game, so he cannot win the second one.

注释：这道题的意思就是输入n个样例，如果 Alex 赢得了第 i 场比赛，则等于 1，如果 Bob 赢得了第 i 场比赛，则等于 2，如果卡尔赢得了第 i 场比赛，则等于 3。看所给的样例输入是否符合比赛情况（是否合理），若合理则“yes”，否则“no”；

我们要分好每次象棋比赛中的胜利者，失败者和裁判是谁。

代码有注释，看下c++代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[105];
int win(int A,int C)//A为赢家，C为裁判，输出输家
{
	if(A==1&&C==2||A==2&&C==1) return 3;
	if(A==3&&C==2||A==2&&C==3) return 1;
 	if(A==1&&C==3||A==3&&C==1) return 2;
}
int main()
{
	ios::sync_with_stdio(false);//缓冲
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	int flag=1;
	int A,B,C=3;//开始就将C列为裁判
	for(int i=0;i<n;i++)
	{
		if(a[i]==C)//如果a[i]与裁判相照应那必定错误
		{
			flag=0;
			break;
		}
		A=a[i];//记录赢家，赢家的值赋给a[i]
		B=win(A,C);//记录失败者
		C=B;//将这一局的失败者记录为下一局的裁判
		
	}
	if(flag)
		cout<<"YES"<<endl;
	else 
		cout<<"NO"<<endl;
	return 0;
}

有改进的地方也请大佬们多多指出