Appleman and Easy Task

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). Thenn lines follow containing the description of the checkerboard. Each of them containsn 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

3xxoxoxoxx

Output

YES

Input

4xxxoxoxooxoxxxxx

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;}