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, ..., dk
acycle if and only if it meets the following condition:

These k dots are different: if
i ≠ j then di is different fromdj.
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 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 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).

 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;
}