Fox and the Dots

B-fox and both DotsTime limit:2000MS Memory Limit:262144KB 64bit IO Format:%i64d &%I 64u Submit Status

Description

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 D_{1, D2, ..., dK a cycle if an D only if it meets the following condition:}

2. These k Dots is different:if i ≠ J and d_{i is Different from DJ.}
4. K is at least 4.
6. All dots belong to the same color.
8. For all 1≤ i ≤ k -1: d_{i 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): t He number of rows and columns of the board.

Then n lines follow, each line contains a string consisting of m characters, expressing colors O f dots in each line. Each character is a uppercase Latin letter.

Output

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

Sample Input

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

Hint

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).

Euler circuit, with BFS made a pass, found a good mess, learn the great God Dfs, listen to Fang said there is a very simple way to continue to look at tomorrow.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5#include <queue>6 using namespacestd;7 Charmaze[ -][ -];8 intvis[ -][ -];9 intn,m;Ten intFlag =0; One intto[4][2] = {{1,0},{-1,0},{0,1},{0,-1}}; A BOOLCheckintXinty) { -     if(x<0|| x>=n| | y<0|| Y&GT;=M)return false; -     //if (Vis[x][y]) return false; the     return true; - } - voidDfsintXintYintPrexintprey) { -     if(!check (x, y))return; +Vis[x][y] =1; -     intpostx,posty; +      for(inti =0; i<4; i++){ APOSTX = x + to[i][0]; atPosty = y + to[i][1]; -         if(Check (postx,posty) &&maze[postx][posty] = = maze[x][y]&& (postx!=prex| | posty!=prey)) { -             if(Vis[postx][posty]) { -Flag =1; -                 return; -             } in DFS (postx,posty,x,y); -         } to     } + } - voidinput () { the  *scanf"%d%d",&n,&m); $      for(inti =0; i<n; i++) scanf ("%s", Maze[i]);Panax Notoginseng      for(inti =0; i<n; i++){ -          for(intj =0; j<m; J + +){ the             if(!Vis[i][j]) { +DFS (i,j,-1,-1); A             } the         } +     } -     if(flag) printf ("yes\n"); $     Elseprintf"no\n"); $ } - intMain () - { the input (); -     return 0;Wuyi}

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Fox and the Dots