Codeforces Round #290 (Div. 2) B. Fox And Two Dots dfs

Description

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, ..., d_k a cycle if and only if it meets the following condition:

1. These k dots are different: if i ≠ j then d_i is different from d_j.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: d_i and d_i + 1 are adjacent. Also, d_k 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.

Sample Input

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

Hint

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).

简单来说就是找一个环  这个环由相同的字母构成    用dfs从每个点出发看能不能再回到初始点

为了防止  出现下面的情况加入了fx fy

AAA

ABA

AAA

(0,0)—>(0,1)—>(0,0)这样没有形成环就直接结束了  加入fx fy是为了防止回头查找  也就是只能让查找不能回头

#include<cstdio>
#include<cstring>
using namespace std;
char map[101][101];
int vis[101][101];
int bg,flog;
int n,m;
int dx[4]={0,-1,1,0};
int dy[4]={-1,0,0,1};
void dfs(int x,int y,int fx,int fy,char s) 
{
	 if(flog==1)
	 {
	   return ;
	 }
		    for(int i=0;i<4;i++)
		  {    	
				     int nx=x+dx[i];
		              int ny=y+dy[i];
		             	if(nx>=0&&ny>=0&&nx<n&&ny<m&&map[nx][ny]==s)
		         {   
			        if(nx==fx&&ny==fy) 
			        {
			 	       continue;
			        }
		           if(vis[nx][ny]==1&&map[nx][ny]==s)
		            {
		            flog=1;
		            return ;
	      	        }
					vis[nx][ny]=1; 
		          dfs(nx,ny,x,y,s);  
	        	}
	     	}
}
int main()
{  
  memset(vis,0,sizeof(vis));
  scanf("%d%d",&n,&m);
   for(int i=0;i<n;i++)
   {
   	scanf("%s",map[i]);
   }
       flog=0;
   for(int i=0;i<n;i++)
   {
   	for(int j=0;j<m;j++)
   	{
   		if(vis[i][j]==0)
   		{
			   	 vis[i][j]=1;
   		    dfs(i,j,-2,-2,map[i][j]);
   		    if(flog==1)
   		    {
   		    	    break;
			   }
		   }
	   }
	   if(flog==1)
	   {
	   	break;
	   }
   }
   if(flog==1)
   {
   	printf("Yes
");
   }
   else
   {
   	   printf("No
");
   }
	return 0;
}