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)

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

#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define MIN(a) *min_element(all(a))
#define MAX(a) *max_element(all(a))
#define SUM(a) accumulate(all(a), 0LL)
#define REP(i, n) for(int (i)=0; (i)<(n); (i)++)
#define RREP(i, n) for(int (i)=(n)-1; (i)>=0; (i)--)
#define FOR(i, m, n) for(int (i)=(m); (i)<(n); i++)
#define FORR(i, m, n) for(int (i)=(n)-1; (i)>=(m); i--)
#define debug(x) cerr << #x << " = " << x << endl
//#define int long long

typedef long long ll;
//ll const MOD = 998244353;
ll const MOD = 1e9+7;
int const inf = 1e9;
ll const INF = 1e18;

inline void ios_(){cin.tie(0); ios::sync_with_stdio(false);}
template<typename T> int size(const T& a){return (int)a.size();}
template<typename T> T Div(T a, T b){return (a + b - 1) / b;}
template<typename T> bool chmin(T& a, const T& b){if(a > b){a = b; return true;} return false;}
template<typename T> bool chmax(T& a, const T& b){if(a < b){a = b; return true;} return false;}

//Graph in adjcency list
struct Graph{
    struct Edge{
        int from;
        int to;
        ll cost;
        Edge(int a, ll b) : to(a), cost(b) {}
        Edge(int a, int b, ll c) : from(a), to(b), cost(c) {}
    };

    int N;  //頂点数
    vector<vector<Edge>> edges;   //隣接リスト
    vector<ll> dist;   //ノードの深さ
    ll diam;            //木の直径
    vector<int> deg;    //入次数
    vector<int> cycle;  //サイクル

    Graph(int n) : N(n), edges(n), dist(n, INF), diam(INF), deg(n, 0){}

    void setEdge(int a, int b, ll c){   //辺を追加
        edges[a].push_back(Edge(a, b, c));
        deg[b]++;
    }

    bool visited(int v){
        return dist[v] < INF;
    }

    void resetDist(){
        for(int i=0; i<N; i++){
            dist[i] = INF;
        }
    }

    void dfs(int p, int v, ll d){    //深さ優先探索
        if(visited(v)) return;

        dist[v] = d;

        for(auto e : edges[v]){
            if(e.to == p) continue;
            dfs(v, e.to, d + e.cost);
        }
    }

    void bfs(int v){        //幅優先探索
        queue<Edge> que;
        for(auto e : edges[v]){
            que.push(e);
            dist[e.to] = e.cost;
        }

        dist[v] = 0;

        while(!que.empty()){
            int u = que.front().to;
            ll d = que.front().cost;
            que.pop();

            for(auto e : edges[u]){
                if(!visited(e.to)){
                    que.push(Edge(e.to, d + e.cost));
                    dist[e.to] = d + e.cost;
                }
            }
        }
    }

    ll diameter(){  //木の直径を求める
        if(diam != INF) return diam;  //計算済みの場合

        dfs(-1, 0, 0);  //頂点0からの最遠頂点vを求める
        int v = max_element(all(dist)) - dist.begin();

        resetDist();

        dfs(-1, v, 0);  //vからの最遠頂点を求める

        return diam = MAX(dist);   //この2頂点間の距離が直径
    }

    void dijkstra(int start){   //ダイクストラ法
        dist[start] = 0;

        auto fn = [](Edge a, Edge b){return a.cost > b.cost;};
        priority_queue<Edge, vector<Edge>, decltype(fn)> que(fn);

        for(auto e : edges[start]) que.push(e);

        while(!que.empty()){
            auto v = que.top().to;
            auto d = que.top().cost;
            que.pop();

            if(visited(v)) continue;
            else dist[v] = d;
            
            for(auto e : edges[v]){
                if(dist[e.to] == INF){
                    que.push(Edge(e.to, e.cost + d));
                }
            }
        }
    }

    bool BellmanFord(int start){    //ベルマンフォード法 (負の閉路がある場合はtrueを返す)
        dist[start] = 0;
        REP(i, N){
            REP(j, N){
                for(auto k : edges[j]){
                    if(visited(j) && dist[j] + k.cost < dist[k.to]){
                        if(i == N-1) return true;  //N回目にも更新があるなら負の閉路が存在
                        dist[k.to] = dist[j] + k.cost;
                    }
                }
            }
        }
        return false;
    }

    vector<int> topologicalSort(){ //トポロジカルソート
        queue<int> que;
        REP(i, N){
            if(deg[i] == 0) que.push(i);
        }

        vector<int> result;

        while(!que.empty()){
            auto v = que.front();
            que.pop();
            result.push_back(v);
            for(auto e : edges[v]){
                deg[e.to]--;
                if(deg[e.to] == 0) que.push(e.to);
            }
        }
        
        return result;
    }

    void findCycle(int p, int v) {
        static vector<bool> seen(N, false), finished(N, false);
        static int pos = -1;

        seen[v] = true;
        cycle.push_back(v);

        for(auto e : edges[v]){
            if(e.to == p || finished[e.to]) continue;
            if(seen[e.to] && !finished[e.to]){
                pos = e.to;
                return;
            }
            findCycle(v, e.to);

            if(pos != -1) return;
        }

        cycle.pop_back();
        finished[v] = true;
    }
};

signed main(){
    int h, w;
    cin >> h >> w;

    vector<vector<char>> grid(h, vector<char>(w));
    REP(i, h){
        REP(j, w){
            cin >> grid[i][j];
        }
    }

    int n = h*w;

    Graph G(n);

    vector<int> dx = {1, 0, -1, 0, 1, 1, -1, -1}, dy = {0, 1, 0, -1, 1, -1, 1, -1};

    REP(i, h){
        REP(j, w){
            if(grid[i][j] == '#') continue;
            REP(k, 4){
                int xi = i + dx[k], yi = j + dy[k];
                if(xi < 0 || xi >= h || yi < 0 || yi >= w){
                    continue;
                }
                if(grid[xi][yi] == '.'){
                    G.setEdge(i*w + j, xi*w + yi, 1);
                }
            }
        }
    }

    ll ans = 0;

    REP(i, n){
        G.bfs(i);
        REP(j, n){
            if(G.dist[j] < INF){
                chmax(ans, G.dist[j]);
            }
        }
        G.resetDist();
    }

    cout << ans << endl;
    return 0;
}