Fox And Two Dots

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本文介绍了一款名为TwoDots的游戏中的算法挑战，即如何在给定的游戏板上找到相同颜色的点构成的环形路径。通过深度优先搜索的方法实现这一目标，并给出了具体的实现代码。

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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:

$\text{[math]}$

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:

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

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).

解析：想要形成一个环必须能够从起点经过深度搜索回到起点，并且排出AA这种非环的情况。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int M = 55;
int fx[4]={-1,1,0,0},fy[4]={0,0,-1,1};
int n,m,ok,vis[M][M],sx,sy;
char s[M][M];
void dfs(int x,int y,int flag)
{
	if(ok) return ;
	vis[x][y]=1;
	for(int i=0;i<4;i++){
		int xx=x+fx[i],yy=y+fy[i];
		if(xx==sx&&yy==sy&&flag>1){    //此处flag大于1或2都可以。 
			ok=1;return ;
		}
		if(xx>=0&&yy>=0&&xx<n&&yy<m&&!vis[xx][yy]&&s[xx][yy]==s[x][y]){
			dfs(xx,yy,flag+1);
        }
    }
}
int main()
{
	while(~scanf("%d%d",&n,&m)){
		ok=0;
		for(int i=0;i<n;i++)
		scanf("%s",s[i]);
		for(int i=0;i<n&&!ok;i++)
		for(int j=0;j<m&&!ok;j++){
			memset(vis,0,sizeof(vis));
			sx=i;sy=j;
			dfs(i,j,0);
		}
		if(!ok) puts("No");
		else puts("Yes");
	}
}