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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

// template version 1.15
using namespace std;
#include <bits/stdc++.h>

// varibable settings
#define int long long
const int INF=1e18;

// define basic macro {{{
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define _rrep(i,n) rrepi(i,0,n)
#define rrepi(i,a,b) for(int i=(int)((b)-1);i>=(int)(a);--i)
#define rrep(...) _overload3(__VA_ARGS__,rrepi,_rrep,)(__VA_ARGS__)
#define each(i,a) for (auto&& i : a)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define mt(a, b, c) make_tuple(a, b, c)
#define divceil(a,b) ((a)+(b)-1)/(b)
#define is_in(x, a, b) ((a)<=(x) && (x)<(b))
#define uni(x) sort(all(x));x.erase(unique(all(x)),x.end())
#define ub upper_bound
#define lb lower_bound
#define posl(A, x) (lower_bound(all(A), x)-A.begin())
#define posu(A, x) (upper_bound(all(A),x)-A.begin())
template<class T> inline void chmax(T &a, const T &b) { if((a) < (b)) (a) = (b); }
template<class T> inline void chmin(T &a, const T &b) { if((a) > (b)) (a) = (b); }

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef long double ld;
typedef pair<int,int> pii;
typedef tuple<int,int,int> iii;

template<typename T> using PQ = priority_queue<T, vector<T>, greater<T>>;
struct Fast { Fast(){ std::cin.tie(0); ios::sync_with_stdio(false); } } fast;

#if defined(PCM) || defined(LOCAL)
  // #include "lib/dump.hpp"
#else
  #define dump(...) 42
  #define dump_1d(...) 42
  #define dump_2d(...) 42
#endif
//}}}

using Cost = int;  // TODO: update
Cost zerocost = 0LL;  // TODO: update
Cost infcost = INF;  // TODO: update
using Pos = int;  // TODO: update
using EDGE = pair<Pos, Cost>;

struct Graph{
  map<Pos, vector<EDGE>> all_edges;
  struct Dist {
    map<Pos, Cost> data;
    Cost operator[](Pos pos){
      if (data.find(pos)!=data.end()) return data[pos];
      else                      return infcost;
    }
  };

  void add_edge(Pos from, Pos to, Cost cost){
    all_edges[from].emplace_back(mp(to, cost));
  }

  auto operator[](Pos pos){
    return all_edges[pos];
  }

  friend auto& operator<<(auto &os, const Graph& G){
    os << G.all_edges;
    return os;
  }

  Dist dijkstra(Pos start){
    Dist d;  // 最短距離
    PQ<pair<Cost, Pos>> pq;
    pq.push(mp(zerocost, start));
    while (!pq.empty()){
      auto cp = pq.top(); pq.pop();
      auto cost = cp.first;
      auto pos = cp.second;
      if (cost < d[pos]) {
        d.data[pos] = cost;
        for (const auto &el: all_edges[pos]){
          int npos = el.first;
          int ncost = cost+el.second;
          if (ncost < d[npos]) pq.push(mp(ncost, npos));
        }
      }
    }
    return d;
  }
};


signed main() {
  int h,w;cin>>h>>w;
  int n = h*w;
  vvi block(h, vi(w));
  rep(i, h){
    string s;cin>>s;
    rep(j, w){
      block[i][j] = (s[j]=='.'?0:1);
    }
  }

  vvi g(n);
  Graph G;

  int di[]={1, -1, 0, 0};
  int dj[]={0, 0, 1, -1};
  rep(i, h)rep(j, w){
    if (block[i][j]) continue;
    int u = i*w+j;

    rep(k, 4){
      int ni = i+di[k];
      int nj = j+dj[k];
      int v = ni*w + nj;
      if (is_in(ni, 0, h) && is_in(nj, 0, w) && !block[ni][nj]){
        g[u].pb(v); // 自分から生える辺だけでよい。そうしないと二重辺になってしまう。
        G.add_edge(u, v, 1);
      }
    }
  }
  
  int ans = 1;
  rep(i, h)rep(j, w){
    if (block[i][j]) continue;

    int u = i*w+j;
    auto d = G.dijkstra(u);
    rep(i, n) chmax(ans, (d[i]==INF ? 1 : d[i]));
  }
  cout << ans << endl;

  return 0;
}