Educational Codeforces Round 33 (Rated for Div. 2) A题. Chess For Three

A. Chess For Three

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 a_i (1 ≤ a_i ≤ 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.

Examples

Input

3
1
1
2

Output

YES

Input

2
1
2

Output

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.

题意：三个人 比赛，两个人下棋 一个人旁观，下棋之后输的人去旁观，胜的人和之前旁观的人比赛下棋 ，

　　　第 1 人 和 第 2 人 先比 ，第 3 人 旁观 ，每组样例先输入一个 n 表示有 n 场比赛（每场比赛必分胜负），

          之后 n 个数 表示 每场比赛胜利的人 ， 检查整个过程  判断每场胜利的人是否 是 正在下棋的 人 中的一个

思路：直接模拟下棋过程 ， 遇到不符合情况的 判错。

#include<stdio.h>
#include<string.h>
#include <iostream>
#include<algorithm>

using namespace std;
#define maxn 200
int num[maxn] ; 

int main(){
    
    int n ; 
    int num1 , num2 , pang ; 
    
    while(~scanf("%d" , &n)){
        for(int i=1 ; i<= n ;i++){
            scanf("%d" , &num[i]) ; 
        }
        num1 = 1 ; // num1  num2 时正在下棋的两个人 
        num2 = 2 ; 
        pang = 3 ; // 旁观的 一个人 
        int temp ;  
        bool flag = true ; 
        for(int i=1 ; i<=n ; i++){
            if(num[i] == num1 ){
                temp = num2 ; 
                num2 = pang ; 
                pang = temp ; 
            } else if(num[i] == num2){
                temp = num1 ; 
                num1 = pang ; 
                pang = temp ; 
            } else {
                flag = false ; 
                break ; 
            }
        }
        if(flag){
            printf("YES
") ; 
        } else printf("NO
") ; 
    }
    
    return 0 ; 
}