B. Fox And Two Dots

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:

$\text{[math]}$

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:

These k dots are different: if i ≠ j then di is different from dj.
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 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 test(s)

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).

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <string>
 7 #include <vector>
 8 #include <set>
 9 #include <map>
10 #include <queue>
11 #include <stack>
12 using namespace std;
13 const int INF = 0x7fffffff;
14 const double EXP = 1e-8;
15 const int MS = 55;
16 int n, m, cnt;
17 
18 char cell[MS][MS];
19 int vis[MS][MS];
20 int dir[4][2] = { 0, 1, 1, 0, 0, -1, -1, 0 };
21 
22 bool dfs(char c, int sx, int sy, int x, int y)
23 {
24     if (sx == x&&sy == y&&vis[sx][sy])
25     {
26         return cnt >= 4 ? true : false;
27     }
28     vis[x][y] = 1;
29     for (int i = 0; i < 4; i++)
30     {
31         int tx = x + dir[i][0];
32         int ty = y + dir[i][1];
33         if (tx >= 0 && tx < n&&ty >= 0 && ty < m&&cell[tx][ty] == c&&vis[tx][ty] == 0 || (tx == sx&&ty == sy))
34         {
35             cnt++;
36             if (dfs(c, sx, sy, tx, ty))
37                 return true;
38             cnt--;
39         }
40     }
41     return false;
42 }
43 
44 
45 bool solve()
46 {
47     for (int i = 0; i < n; i++)
48     {
49         for (int j = 0; j < m; j++)
50         {
51             memset(vis, 0, sizeof(vis));
52             cnt = 0;
53             if (dfs(cell[i][j], i, j, i, j))
54                 return true;
55         }
56     }
57     return false;
58 }
59 int main()
60 {
61     cin >> n >> m;
62     for (int i = 0; i < n; i++)
63         cin >> cell[i];
64     if (solve())
65         cout << "Yes" << endl;
66     else
67         cout << "No" << endl;
68     return 0;
69 }