CF 510b Fox and both Dots

Fox Ciel is playing a mobile puzzle game called ". The basic levels is played on a board of size n x m cells, like this:

Each cell contains a dot this has some color. We'll use the different uppercase Latin characters to express different colors.

The key of this game was 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:

2. These k Dots is different:if i ≠ J and di is different fro M DJ.
4. K is at least 4.
6. All dots belong to the same color.
8. For all 1≤ i ≤ k -1: di and di + 1 is Adjac Ent. Also, dK and D1 should Also be adjacent. Cells x and y is called adjacent if they share an edge.

Determine if there exists a cycle on the field.

Input

The first line contains integers n and m (2≤ n, m ≤50): the number O f 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 a uppercase Latin letter.

Output

Output "Yes" if there exists a cycle, and "No" otherwise.

Examplesinput

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).

The main is to use DFS to determine whether to return to the point of departure, and ensure that no backtrack (so the DFS function defines four variables)

#include <cstdio>#include<cstring>intflag[ -][ -];Charstr[ -][ -];intn,m,k;intdx[4]={-1,1,0,0};intdy[4]={0,0,-1,1};voidDfsintXintYintCxintcy) {    if(Flag[x][y] = =1) {k=1; return ; } Flag[x][y]=1; intI,nx,ny;  for(i =0; I <4; i++) {NX=x+Dx[i]; NY=y+Dy[i]; if(Str[nx][ny] = = Str[x][y] && (NX! = CX | | NY! =cy))                    {DFS (nx,ny,x,y); }    }    }intMain () { while(SCANF ("%d%d", &n,&m)! =EOF) {k=0; inti,j; memset (Flag,0,sizeof(flag));  for(i =1; I <= N; i++) {scanf ("%s", str[i]+1); }         for(i =1; I <= N; i++)        {             for(j =1; J <= M; J + +)            {                if(Flag[i][j] = =1)                                    Continue;                                DFS (I,J,I,J); if(k = =1)                                     Break; }            if(k = =1)                 Break; }         if(k = =1) printf ("yes\n"); Elseprintf ("no\n"); }}

CF 510b Fox and both Dots