Fox And Two Dots 迷宫 深搜

Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:

Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.

The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:

1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.

Determine if there exists a cycle on the field.

Input

The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.

Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.

Output

Output "Yes" if there exists a cycle, and "No" otherwise.

Example

Input

3 4
AAAA
ABCA
AAAA

Output

Yes

Input

3 4
AAAA
ABCA
AADA

Output

No

Input

4 4
YYYR
BYBY
BBBY
BBBY

Output

Yes

Input

7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB

Output

Yes

Input

2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ

Output

No

Note

In first sample test all 'A' form a cycle.

In second sample there is no such cycle.

The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).

 

 

 

 

 

 

#include<stdio.h>
#include<string.h>
int m,n;
int dis[8]={1,0,-1,0,0,1,0,-1};//　 方向数组　　四个方向
int vis[100][100];
char ch[100][100];
int flag;
void dfs(int x,int y,int fx,int fy)//此处的四个变量代表当前点的坐标和上一個走过的点的坐标
{
    flag=0;
    int i,j;
    int tx,ty;
    for(i=0;i<8;i+=2)//四个方向都要遍历
    {
        tx=x+dis[i];
        ty=y+dis[i+1];//将要走的点的坐标
        if(tx<0||ty<0||tx>=m||ty>n-1||ch[tx][ty]!=ch[x][y]||tx==fx&&ty==fy)//这些点要满足四个条件　数组不越接界　和与上一个字母相同　没有走回去
            continue;

            if(vis[tx][ty])//只要找到一个被标记的就成环
            {
                flag=1;
                return ;
            }

          
            vis[tx][ty]=1;
            dfs(tx,ty,x,y);//此处的三四个变量必须是x,y　不能改变fx,fy的值


    }
}
int main()
{
    int i,j;
    while(scanf("%d %d",&m,&n)!=EOF)
    {
        for(i=0;i<m;i++)
        {
            scanf("%s",ch[i]);//此处特别注意　若用％Ｃ输入　加　　getchar(); 吸收回车
        }

        for(i=0;i<m&&!flag;i++)//因为每一个点都有可能成为迷宫的起点所以要遍历每一个点
        {
            for(j=0;j<n;j++)
            {
                memset(vis,0,sizeof(vis));//清空数组
                vis[i][j]=1;//将选出来的点标记上
                dfs(i,j,i,j);//进入深搜
                if(flag)
                    break;
            }
        }
        if(flag)
            printf("Yes\n");
        else
            printf("No\n");

    }


    return 0;
}