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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef vector<char> VC;
typedef vector<double> VD;
typedef vector<string> VS;
typedef vector<ll> VLL;
typedef vector<PLL> VP;
const static int INF = 1000000000;
const static int MOD = 1000000007;
const int dx[4]={1, 0, -1, 0};
const int dy[4]={0, 1, 0, -1};
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
#define rep(i,n) for (ll i=0; i<(ll)(n); i++)
#define repd(i,n) for (ll i=n-1; i>=0; i--)
#define rept(i,m,n) for(ll i=m; i<n; i++)
#define stl_rep(itr, x) for (auto itr = x.begin(); itr != x.end(); ++itr)
#define all(x) (x).begin(), (x).end()
#define F first
#define S second
#define PF push_front
#define PB push_back
#define SORT(V) sort((V).begin(), (V).end())
#define RVERSE(V) reverse((V).begin(), (V).end())
#define paired make_pair
#define PRINT(V) for(auto v : (V)) cout << v << " "
//charを整数に
int ctoi(char c) { if (c >= '0' && c <= '9') { return c - '0'; } return 0; }
// 累積和 for (int i = 0; i < N; ++i) s[i+1] = s[i] + a[i]; 
void cum_sum(int N,vector<double> a, vector<double> &s){ for(int i=0; i<N; i++){ s[i+1]=s[i]+a[i];}}
//ユークリッドの控除法
ll gcd(ll a,ll b){
    if(b == 0) return a;
    return gcd(b,a%b);
}
//最小公倍数
ll lcm(ll a,ll b){
    ll g = gcd(a,b);
    return a / g * b; // Be careful not to overflow
}
//素数判定
bool is_prime(long long n) {
    if (n <= 1) return false;
    for (long long p = 2; p * p <= n; ++p) {
        if (n % p == 0) return false;
    }
    return true;
}
int getdigit(ll num){
    unsigned digit=0;
    while(num!=0){
        num /= 10;
        digit++;
    }
    return digit;
}
//空白文字も入力 getline(cin, S);
//桁数指定 setprecision

int main()
{
    ios::sync_with_stdio(0); 
    cin.tie(0); 
    cout.tie(0);
    int H, W;
    cin >> H >> W;
    vector<string> S(H);
    rep(i,H){
        cin >> S[i];
    }
    ll max_dis=0;
    for(int i=0; i<H; i++){
        for(int j=0; j<S[i].size(); j++){
            if(S[i].at(j)=='.'){
                int i_x=i;
                int i_y=j;
                queue<pair<int, int>> que;
                que.push(make_pair(i_x, i_y));
                vector<vector<ll>> dist(H, vector<ll>(W));
                dist[i_x][i_y]=0;
                while(!que.empty()){
                    pair<int, int> fir;
                    fir=que.front();
                    que.pop();
                    int if_x=fir.F;
                    int if_y=fir.S;
                    for(int k=0; k<4; k++){
                        int i2_x=if_x+dx[k];
                        int i2_y=if_y+dy[k];
                        if(i2_x>=0 && i2_x<H && i2_y>=0 && i2_y<W && S[i2_x][i2_y]!='#'){
                            if(dist[i2_x][i2_y]!=0) continue;
                            que.push(make_pair(i2_x, i2_y));
                            dist[i2_x][i2_y]=dist[if_x][if_y]+1;
                            max_dis=max(max_dis,dist[i2_x][i2_y]);
                        }                        
                    }
                }
            }
        }
    }
    cout << max_dis << endl;
}