Fox And Two Dots

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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:



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.

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");
}
}