B. Fox And Two Dots

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

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).

 

题意：有多种颜色，问你颜色相同的点能否组成一个环（只能上下左右的走）

 

本题用DFS，不同的是每次递归，要记录上一个走过点点，保证下次不能再走在个点，这样一直走，如果能走到标记过的点，那么就说明存在回路。

 

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int px[4]={0,0,1,-1};//方向 int py[4]={-1,1,0,0}; int vis[55][55];//标记 char map[55][55]; int n,m,ans; void DFS(int x,int y,int cx,int cy)//cx，cy记录上次走过的点 { char c; c=map[x][y]; for(int i=0;i<4;i++) { int nx=x+px[i]; int ny=y+py[i]; if(map[nx][ny]==c&&nx>=0&&nx<n&&ny>=0&&ny<m) { if(nx==cx&&ny==cy) //不走上次走过的点 { continue; } if(vis[nx][ny]) //如果它被标记，就说明存在回路 { ans=1; return ; } vis[nx][ny]=1; DFS(nx,ny,x,y); } } return ; } int main() { while(scanf("%d%d",&n,&m)!=EOF) { ans=0; for(int i=0;i<n;i++) scanf("%s",map[i]); memset(vis,0,sizeof(vis)); for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { if(!vis[i][j]) { vis[i][j]=1; DFS(i,j,i,j); } if(ans) break; } if(ans) break; } if(ans) printf("Yes\n"); else printf("No\n"); } return 0; }