def C_BFS(R, C, sy, sx, gy, gx, c):
    """
    R:行数
    C:列数
    sy,sx:スタートの座標(sy行sx列)
    gy,gx:ゴールの座標(gy行gx列)
    c:迷路の形状(.:通路,#:壁)
    """
    INF = float('inf')
    sy -= 1
    sx -= 1
    gy -= 1
    gx -= 1
 
    def bfs(maze, row, column, sy, sx, gy, gx):
        # 原点は(0,0)とする(問題の制約によっては原点をずらす必要がある)
        queue = []
        queue.insert(0, (sy, sx))
        dist = [[INF for _ in range(column)] for _ in range(row)]
        dist[sy][sx] = 0  # yが行数、xが列数を意味するときはこの順に書く
        while len(queue):
            y, x = queue.pop()
            if x == gx and y == gy:
                # ゴール座標と一致するなら
                break
            for i in range(4):
                nx, ny = x + [1, 0, -1, 0][i], y + \
                    [0, 1, 0, -1][i]  # (x,y)の右、上、左、下
                if 0 <= nx and nx < column and 0 <= ny and ny < row \
                        and maze[ny][nx] != '#' and dist[ny][nx] == INF:
                    #(nx,ny)が迷路内であり、壁でなく、まだ距離が確定していないとき
                    queue.insert(0, (ny, nx))  # 次に探索する座標
                    dist[ny][nx] = dist[y][x] + 1
        return dist[gy][gx]
 
    maze = [['' for _ in range(C)] for _ in range(R)]
    for i, tmp in enumerate(c):
        for j in range(C):
            maze[i][j] = tmp[j]
 
    ans = bfs(maze, R, C, sy, sx, gy, gx)
    return ans

R , C = [int(i) for i in input().split()]
c = [input() for _ in range(R)]

ans = 0
for sy in range(1, R+1):
    for sx in range(1, C+1):
        if c[sy-1][sx-1] != '#':
            for gy in range(1, R+1):
                for gx in range(1, C+1):
                    if c[gy-1][gx-1] != '#':
                        count_tmp = C_BFS(R, C, sy, sx, gy, gx, c)
                        if ans < count_tmp:
                            ans = count_tmp
print(ans)