CodeForces 510B B. 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:

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

3 4AAAAABCAAAAA

output

Yes

input

3 4AAAAABCAAADA

output

No

input

4 4YYYRBYBYBBBYBBBY

output

Yes

input

7 6AAAAABABBBABABAAABABABBBABAAABABBBABAAAAAB

output

Yes

input

2 13ABCDEFGHIJKLMNOPQRSTUVWXYZ

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).

问相同的字母能不能组成一个环（DFS）

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char map[60][60];int vis[60][60];int n,m,i,j,flag;int dir[4][2]={1,0,0,1,-1,0,0,-1};void dfs(int x,int y,int parx,int pary,char s){//vis[x][y]=1;int l;if(flag==1)return ;for(l=0;l<4;l++){int fx=x+dir[l][0];int fy=y+dir[l][1];if(fx>=0&&fx<n&&fy>=0&&fy<m&&map[fx][fy]==s){if(fx==parx&&fy==pary)//看这次走到的点是不是上一次所在的点，如果是则直接开始下一个循环 continue;if(vis[fx][fy]==1)//这里遇到了已经标记过了的点，则说明构成了环. {flag=1;return;}vis[fx][fy]=1;dfs(fx,fy,x,y,s);}}}int main(){while(scanf("%d%d",&n,&m)!=EOF){for(i=0;i<n;i++)scanf("%s",map[i]);flag=0;memset(vis,0,sizeof(vis));for(i=0;i<n;i++){for(j=0;j<m;j++){if(vis[i][j]==0){vis[i][j]=1;dfs(i,j,-1,-1,map[i][j]);}if(flag==1)break;}if(flag==1)break;}if(flag==1)printf("Yes\n");elseprintf("No\n");}}