Codeforces Round #263 (Div. 2) A. Appleman and Easy Task【地图型搜索/判断一个点四周‘o’的个数的奇偶】

A. Appleman and Easy Task

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

3
xxo
xox
oxx

output

YES

input

4
xxxo
xoxo
oxox
xxxx

output

NO

【题意】：Two cells of the board are adjacent if they share a side. 指的是四周

【Code】：

#include <bits/stdc++.h>
using namespace std;
const int N =110;
int cnt,n;

char a[N][N];
int dir[][2]={ {-1,0},{0,-1},{1,0},{0,1} };

int dfs(int x,int y)
{
    for(int i=0;i<4;i++)
    {
        int dx=x+dir[i][0];
        int dy=y+dir[i][1];

        if(dx>=0&&dy>=0&&dx<n&&dy<n)
        {
            if(a[dx][dy]=='o')
                cnt++;
        }
    }
    return cnt;
}
bool check()
{
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            if(dfs(i,j)&1)
                return false;
        }
    }
    return true;
}
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)
        scanf("%s",a[i]);

    printf("%s\n",check()?"YES":"NO");
	return 0;
}