B. Fox And Two Dots（迷宫找环问题）

题目链接：http://codeforces.com/problemset/problem/510/B

B. Fox And Two Dots

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.

Examples

input

Copy

3 4
AAAA
ABCA
AAAA

output

Copy

Yes

input

Copy

3 4
AAAA
ABCA
AADA

output

Copy

No

input

Copy

4 4
YYYR
BYBY
BBBY
BBBY

output

Copy

Yes

input

Copy

7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB

output

Copy

Yes

input

Copy

2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ

output

Copy

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).

题意：给出一个迷宫，求能不能找到一个环；

那次比赛差一丢丢丢丢丢丢.......（还是太菜）

#include
#include
#include 
using namespace std;
int n,m;
char vec[55][55];
int vis[55][55];
int flag=0;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
void dfs(int bx,int by,int  fx,int fy,char ch){
	if(flag){
		return ;
	}
	for(int i=0;i<4;i++){
		int x= bx + dir[i][0];
		int y= by + dir[i][1];
		if(x>=0 && x=0 && y>n>>m;
	memset(vis,0,sizeof(vis));
	for(int i=0;i