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:

These k dots are different: if i ≠ j then di is different from dj.
k is at least 4.
All dots belong to the same color.
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.

Examples
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).`
题目的意思给你一个图，不同的字母代表不同的颜色，问是否存在一个由相同的颜色形成的一个环，这个环的节点最少是四个

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
char s[55][55];
bool vis[55][55];
int M,N,step,flag;//改了WA WA了改 改改WA WA 再改改 
int bx,by; 
int dir[4][2]={1,0,0,1,-1,0,0,-1};//下右上左 
void DFS(int x,int y,int prex,int prey)//x,y,是当前节点的坐标,prex,prey是上一个节点的坐标
{
	if(flag)//说明已经找到
	return ;
	for(int i=0;i<4;i++)
	{
		int X,Y;
		X=x+dir[i][0];
		Y=y+dir[i][1];
		if(X>=0&&Y>=0&&X<M&&Y<N&&s[X][Y]==s[bx][by])
		{
			if(X==prex&&Y==prey)//不能在走回上一个节点
			continue;
			if(vis[X][Y])//我这判断能不能形成环 
			{
				flag=1;
				return ;
			}
			if(!vis[X][Y])
			{
				vis[X][Y]=1;
				DFS(X,Y,x,y);//这到题而言不能回退，设置prex,prey也是这个目的
			}
		}
	}
	return ;
}
int main()
{
	while(~scanf("%d%d",&M,&N))
	{
		int i,j;
		flag=0;
		for(i=0;i<M;i++)
		scanf("%s",s[i]);
		
		for(i=0;i<M;i++)
		{
			for(j=0;j<N;j++)
			{
				memset(vis,0,sizeof(vis));
				bx=i;
				by=j;
				vis[i][j]=1;
				DFS(i,j,-1,-1);
				if(flag)
				break;
				
			}
			if(flag)
			break;
		}
		if(flag)
		printf("Yes\n");
		else
		printf("No\n");
	}
	return 0;
}