CodeForces 510B——Fox And Two Dots【DFS】

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 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).

题目大意：给你一个由字母组成的矩阵，问相同的字母能否连接成一个环。

题目思路：用DFS搜索每一个点并标记，DFS传递时注意传入当前节点的前一个点的坐标，并把起始点标记，如果最后又搜索到起始点则成环。

#include
#include
#include
#include
using namespace std;
int n,m;
char mp[55][55];
bool vis1[55][55],vis2[60];
bool flag;
int dx[4]={1,0,-1,0},dy[4]={0,-1,0,1};
//用fx,fy记录上一个点的坐标
void DFS(int x,int y,int fx,int fy,char c){
    //一再次找到当前位置为递归结束的点
    vis1[x][y]=1;
    for(int i=0;i<4;i++){
        int nx=x+dx[i];
        int ny=y+dy[i];
        if(nx>=0&&nx=0&&ny