A - Fox And Two Dots（dfs迷宫）

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 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).

dfs迷宫     看是否成环，代码如下

#include<stdio.h>
#include<string.h>
char a[124][124];//1  0 -1 0 1  1 0  -1 0
int vis[125][125],dis[4][2]={0,-1,  0,1,  1,0,  -1,0},n,m;//dis表示方向 共上下左右四个
int flag=0;

void dfs(int x,int y,int fx,int fy)//x，y为此步的坐标，fx，fy表示上一步
{
    for(int i=0;i<4;i++)
    {
        int tx=x+dis[i][0];
        int ty=y+dis[i][1];//tx，ty表示下一步要走的坐标

        if(tx>=0 && tx<n && ty>=0 && ty<m && a[tx][ty]==a[x][y] && !(ty==fy&&tx==fx) )
        {//不能越界    此步与下步的图案要相同    要走的不能是上一步

            if(vis[tx][ty])//遇到标记   说明已经成环
            {
                flag=1;
                return;
            }
            vis[tx][ty]=1;

            dfs(tx,ty,x,y);
        }
    }
}

int main()
{

    while(~scanf("%d%d",&n,&m))
    {
        int i,j;
        flag=0;
        for(i=0;i<n;i++)
            scanf("%s",a[i]);//如果使用%c输入注意getchar的位置
        memset(vis,0,sizeof(vis));

        for(i=0;i<n;i++)//遍历每个点
        {
            for(j=0;j<m;j++)
            {
                memset(vis,0,sizeof(vis));//每次都要清空以确保下次搜索不会与遇到上次的标记,还有另一种不清空思路是加个if判断点是否被标记
                vis[i][j]=1;
                dfs(i,j,i,j);
                if(flag)
                    break;
            }
            if(flag)
            break;
        }
        if(flag)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}