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: Accepted)

from collections import deque
#探索
def bfs(Sy,Sx):

    dis=[[-1]*W for i in range(H)]
    dis[Sy][Sx]=0
    q=deque()
    q.append([Sy,Sx])#pairとして入れる
    nun=0

    #qの中身が無くなるまで続ける(bfsのメイン部分)
    while(len(q)!=0):
        y=q[0][0]
        x=q[0][1]
        q.popleft()
        
        #四方向それぞれ条件を満たせるか調べていく
        for dy,dx in ([0, 1], [1, 0], [0, -1], [-1, 0]):
            ny=y+dy
            nx=x+dx
            if ny<0 or ny>=H or nx<0 or nx>=W:
                continue
            if field[ny][nx]=="#":
                continue
            if dis[ny][nx]!=-1:
                continue
            dis[ny][nx]=dis[y][x]+1
            nun=max(nun,dis[ny][nx])
            q.append([ny,nx])
    return nun
  
#入力
H,W=map(int,input().split())
field=[input() for i in range(H)]

#各開始地点からの距離をそれぞれ求め最大のものを出力
ans=-1
for y in range(H):
    for x in range(W):
        if field[y][x]=="#":
            continue
        ans=max(ans,bfs(y,x))
print(ans)