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.

Sample 1

3 4
AAAA
ABCA
AAAA

Yes

Sample 2

3 4
AAAA
ABCA
AADA

No

Sample 3

4 4
YYYR
BYBY
BBBY
BBBY

Yes

Sample 4

7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB

Yes

Sample 5

2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ

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).

#include <iostream>
using  namespace std;
#include<string>
#include<vector>
#include<algorithm>
#include<string.h>
#pragma warning (disable:4996)
#include<stack>
#include<set>
bool flag = false;
int n, m;
int book[55][55];
char graph[55][55];
int diry[4] = { 1,-1,0,0 };
int dirx[4] = { 0,0,1,-1 };
void dfs(int x, int y, int frox, int fory, char ch);
int main() {cin >> n >> m;for (int cnt = 0; cnt < n; cnt++)for (int cnt1 = 0; cnt1 < m; cnt1++)cin >> graph[cnt][cnt1];for (int cnt = 0; cnt < n; cnt++) {for (int cnt1 = 0; cnt1 < m; cnt1++) {char temp = graph[cnt][cnt1];if (!book[cnt][cnt1]){dfs(cnt1, cnt, -1, -1, temp);//cnt是行数,cnt1是列数if (flag) {cout << "Yes" << endl;return 0;}}}}cout << "No" << endl;return 0;
}
void dfs(int x, int y, int frox, int froy, char ch) {if (x<0 || y<0 || x>m - 1 || y>n - 1)//n是行数,m是列数return;if (ch != graph[y][x])return;if (book[y][x]) {flag = true;return;}book[y][x] = true;for (int cnt = 0; cnt < 4; cnt++) {int nex = dirx[cnt] + x;int ney = diry[cnt] + y;if (nex == frox && ney == froy)continue;dfs(nex, ney, x, y, ch);}
}

 思路:就是通过深度搜索,只要遍到原点就可以了;

我踩过的坑,我一直卡测试点11,查了好几遍,发现是因为x,y在输入进函数时写反了,然后猜测可能是因为这样会导致某个节点没有扫到,或者说越界扫了某个节点,然后本来那个节点是满足题意的.