B. Fox And Two Dots

B. Fox And Two Dots

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
using namespace std;
const int INF = 0x7fffffff;
const double EXP = 1e-8;
const int MS = 55;
int n, m, cnt;

char cell[MS][MS];
int vis[MS][MS];
int dir[4][2] = { 0, 1, 1, 0, 0, -1, -1, 0 };

bool dfs(char c, int sx, int sy, int x, int y)
{
    if (sx == x&&sy == y&&vis[sx][sy])
    {
        return cnt >= 4 ? true : false;
    }
    vis[x][y] = 1;
    for (int i = 0; i < 4; i++)
    {
        int tx = x + dir[i][0];
        int ty = y + dir[i][1];
        if (tx >= 0 && tx < n&&ty >= 0 && ty < m&&cell[tx][ty] == c&&vis[tx][ty] == 0 || (tx == sx&&ty == sy))
        {
            cnt++;
            if (dfs(c, sx, sy, tx, ty))
                return true;
            cnt--;
        }
    }
    return false;
}


bool solve()
{
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            memset(vis, 0, sizeof(vis));
            cnt = 0;
            if (dfs(cell[i][j], i, j, i, j))
                return true;
        }
    }
    return false;
}
int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i++)
        cin >> cell[i];
    if (solve())
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}