Fox And Two Dots

Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of sizen × 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 dotsd₁, d₂, ..., d_k acycle if and only if it meets the following condition:

1. These k dots are different: if i ≠ j then d_i is different fromd_j.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: d_i and d_i + 1 are adjacent. Also,d_k andd₁ should also be adjacent. Cellsx 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 andm (2 ≤ n, m ≤ 50): the number of rows and columns of the board.

Then n lines follow, each line contains a string consisting ofm 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 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).

 思路：判断是否成环是其难点。注意不能往回走。

判断条件：1、颜色全部相同。

                    2、被标记过。

#include<stdio.h>#include<string.h>char str[1005][1005];int vis[1005][1005];int xn[4]={0,-1,1,0}; //简化寻找点的过程。int yn[4]={1,0,0,-1};int n,m,flag;int judge(int x,int y){    if(x>=0 && x<n && y>=0 && y<m)        return 1;    return 0;}void dfs(int x,int y,int qx,int qy) //qx qy为上一次走过的点的坐标。{    int tempx,tempy,i;    vis[x][y]=1; //标记走过的点    for(i=0;i<4;i++)    {        tempx=x+xn[i]; //寻找下一个点的坐标        tempy=y+yn[i];        if(judge(tempx,tempy) && (tempx!=qx || tempy!=qy) && str[tempx][tempy]==str[x][y])//1、不越界。        {                                                                                 //2、不是前一个点。            if(vis[tempx][tempy])                                                         //3、颜色相同。            {                flag=1;                return;            }            dfs(tempx,tempy,x,y);        }    }    return;}int main(){    int i,j;    while(scanf("%d%d",&n,&m)!=EOF)    {        flag=0;        memset(vis,0,sizeof(vis));        for(i=0;i<n;i++)            scanf("%s",str[i]);        for(i=0;i<n;i++)        {            for(j=0;j<m;j++)            {                if(!vis[i][j])                {                    dfs(i,j,-1,-1);                    if(flag)                        break;                }            }            if(flag)                break;        }        if(flag)            printf("Yes\n");        else            printf("No\n");    }    return 0;}