Fox And Two Dots

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).

代码：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 char a[55][55];
 6 int v[55][55];
 7 int fx[4]={0,0,1,-1},fy[4]={1,-1,0,0};
 8 int n,m,ans,e1,e2;
 9 int dfs(int x,int y,int nol)
10 {
11     if(ans) return ans;
12     v[x][y]=1;
13     for(int i=0;i<4;i++)
14      {int xx,yy;
15       xx=x+fx[i];
16       yy=y+fy[i];
17       if(xx==e1&&yy==e2&&nol>2)     
18         {ans=1;
19          return ans; 
20     
21         }
22       if(xx>=0&&yy>=0&&xx<n&&yy<m&&a[xx][yy]==a[x][y]&&!v[xx][yy])
23         dfs(xx,yy,nol+1);
24            
25      }
26     
27     
28 }
29 int main()
30 {
31     
32     scanf("%d %d",&n,&m);
33     for(int i=0;i<n;i++)
34       scanf("%s",a[i]);
35     
36     ans=0;  
37     for(int i=0;i<n&&!ans;i++)
38       for(int j=0;j<m&&!ans;j++)
39         {
40          memset(v,0,sizeof(v));
41          e1=i;
42          e2=j;
43          dfs(i,j,0);
44            
45                
46                
47         } 
48     if(ans) printf("Yes
");
49     else printf("No
");         
50     return 0;
51 }