Problem:
Score : 400 points

Problem Statement

Takahashi has a maze, which is a grid of H \times W squares with H horizontal rows and W vertical columns.

The square at the i-th row from the top and the j-th column is a "wall" square if S_{ij} is #, and a "road" square if S_{ij} is ..

From a road square, you can move to a horizontally or vertically adjacent road square.

You cannot move out of the maze, move to a wall square, or move diagonally.

Takahashi will choose a starting square and a goal square, which can be any road squares, and give the maze to Aoki.

Aoki will then travel from the starting square to the goal square, in the minimum number of moves required.

In this situation, find the maximum possible number of moves Aoki has to make.

Constraints

1 \leq H,W \leq 20

S_{ij} is . or #.

S contains at least two occurrences of ..

Any road square can be reached from any road square in zero or more moves.

Input

Input is given from Standard Input in the following format:

H W
S_{11}...S_{1W}
:
S_{H1}...S_{HW}

Output

Print the maximum possible number of moves Aoki has to make.

Sample Input 1

3 3
...
...
...

Sample Output 1

4

If Takahashi chooses the top-left square as the starting square and the bottom-right square as the goal square, Aoki has to make four moves.

Sample Input 2

3 5
...#.
.#.#.
.#...

Sample Output 2

10

If Takahashi chooses the bottom-left square as the starting square and the top-right square as the goal square, Aoki has to make ten moves.

Solution in Python (3.4.3) (judge verdict: Wrong Answer)

from collections import deque
 
# (sx, sy) から 全点への最短距離を求める
# 辿り着けないと INF
def bfs(maze, sx, sy, n, m):
    # すべての点を 0 で初期化
    d = [[0] * m for i in range(n)]
 
    # 移動4方向のベクトル
    dx = [1, 0, -1, 0]
    dy = [0, 1, 0, -1]
 
    # スタート地点をキューに入れ、その点の距離を0にする
    que = deque([])
    que.append((sx, sy))
    d[sx][sy] = 0
 
    # キューが空になるまでループ
    while que:
        # キューの先頭を取り出す
        p = que.popleft()
        # 取り出してきた状態がゴールなら探索をやめる
 
        # 移動4方向をループ
        for i in range(4):
            # 移動した後の点を (nx, ny) とする
            nx = p[0] + dx[i]
            ny = p[1] + dy[i]
 
            # 移動が可能かの判定とすでに訪れたことがあるかの判定
            # d[nx][ny] != INF なら訪れたことがある
            if 0 <= nx < n and 0 <= ny < m and maze[nx][ny] != "#" and d[nx][ny] == 0:
                # 移動できるならキューに入れ、その点の距離を p からの距離 +1 で確定する
                que.append((nx, ny))
                d[nx][ny] = d[p[0]][p[1]] + 1
 
    return max(max(d))
 
 
n, m = map(int, input().split())
maze = [list(input()) for i in range(n)]
 
#print(maze)
 
bfs_list = [0]
 
for i in range(n):
  for j in range(m):
    # スタートとゴールの座標を探す
      sx = i
      sy = j
      if maze[sx][sy] == "#":
      	continue
      bfs_list.append(bfs(maze, sx, sy, n, m))
 
print(max(bfs_list))