Appleman and Easy Task

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Toastman 提出了一个关于 n×n 棋盘的任务，棋盘上的每个格子包含字符 'x' 或 'o'，需要判断每个格子是否有偶数个相邻的 'o'。本文提供了一个 C++ 解决方案来解决这个问题。

Description

Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?

Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side.

Input

The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces.

Output

Print "YES" or "NO" (without the quotes) depending on the answer to the problem.

Sample Input

Input

3
xxo
xox
oxx

Output

YES

Input

4
xxxo
xoxo
oxox
xxxx

Output

NO

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
char mp[110][110];
int n;

int main(){
    while(cin>>n){
        int ok=1;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++)
                cin>>mp[i][j];
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                int cnt=0;
                if(mp[i+1][j]=='o')
                    cnt++;
                if(mp[i-1][j]=='o')
                    cnt++;
                if(mp[i][j-1]=='o')
                    cnt++;
                if(mp[i][j+1]=='o')
                    cnt++;
                if(cnt&1){//取cnt的最左边一位.例如cnt是0011,cnt&1得到0001,如果cnt是0110,cnt&1得到0000，即判断奇偶，效率比cnt%2高
                    ok=0;
                    break;
                }
            }
        }
        if(ok)  cout<<"YES"<<endl;
        else   cout<<"NO"<<endl;
    }
    return 0;
}