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: Wrong Answer)

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

/* msi1427 */

#define ll           long long
#define ull          long long unsigned
#define LD           long double
#define FOR(i,n)     for(int i=0;i<n;i++)
#define FLR(i,n)     for(ll i=0;i<n;i++)
#define RFOR(i,n)    for(int i=n-1;i>=0;i--)
#define RFLR(i,n)    for(ll i=n-1;i>=0;i--)
#define rep(i,a,b)   for(int i=a;i<=b;i++)
#define repl(i,a,b)  for(ll i=a;i<=b;i++)
#define casep(z)     cout << "Case " << z << ": ";
#define Fast         ios_base::sync_with_stdio(0);cin.tie(0);
#define sz(a)        a.size()
#define all(x)       x.begin(),x.end()
#define SORT(a,n)    sort(a,a+n)
#define RSORT(a,n)   sort(a,a+n,greater<ll>())
#define pii          pair<int,int>
#define pll          pair<ll,ll>
#define tll          tuple<ll,ll,ll>
#define VSORT(v)     sort(all(v))
#define VRSORT(v)    sort(v.rbegin(),v.rend())
#define pb           push_back
#define eb           emplace_back
#define pf           push_front
#define pob          pop_back()
#define pof          pop_front()
#define ff           first
#define ss           second
#define tt           true
#define fs           false
#define MP           make_pair
#define MEM(x,n)     memset((x),n,sizeof(x))
#define bsrch(a,n,x) binary_search(a,a+n,x)
#define vi           vector<int>
#define vll          vector<ll>
#define PI           acos(-1.0)
#define LB(a,x)      (lower_bound(all(a),x)-a.begin()) //  first element in the range [first,last) which does not compare less than val.
#define UB(a,x)      (upper_bound(all(a),x)-a.begin()) //  first element in the range [first,last) which compares greater than val.
#define PQ           priority_queue
#define heap         PQ< ll , vector<ll> , greater < ll > >
#define REV(x)       reverse(all(x))
#define mpii         map<int,int>
#define mpll         map<ll,ll>
#define cig          cin.ignore()
#define INF          (ll)1e18
#define EPS          1e-9
#define prec(n)      fixed << setprecision(n)
#define lb           lower_bound
#define ub           upper_bound
#define fin          freopen("in.txt","r",stdin)
#define fout         freopen("out.txt","w",stdout)
#define sc           scanf
#define pt           printf
#define np           next_permutation
#define IT(it,x)     for(auto it=x.begin();it!=x.end();it++)
#define sci(x)       scanf("%d",&x);
#define scl(x)       scanf("%lld",&x);
#define scd(x)       scanf("%lf",&x);
#define scs(x)       scanf("%s",x);

#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(min(a,b),min(c,d))
#define max4(a,b,c,d)   max(max(a,b),max(c,d))
#define max5(a,b,c,d,e) max(max3(a,b,c),max(d,e))
#define min5(a,b,c,d,e) min(min3(a,b,c),min(d,e))
#define EulerGamma      0.57721566490153286060651209

typedef
tree<
  ll,
  null_type,
  less<ll>,
  rb_tree_tag,
  tree_order_statistics_node_update>
ordered_set;

ll dx[] = {1,-1,0,0};
ll dy[] = {0,0,1,-1};

ll ceiling(ll x,ll y)
{
    if(x%y) return ((x/y)+1);
    else return (x/y);
}

ll gcd(ll x,ll y)
{
    return ((y==0)?x:gcd(y,x%y));
}

ll lcm(ll x,ll y)
{
    return ((x*y)/gcd(x,y));
}

ll BMod(ll B,ll P,ll M)
{
    ll R=1;
    B=B%M;
    while(P)
    {
        if(P&1)
            R=(R*B)%M;
        P >>= 1;
        B=(B*B)%M;
    }
    return R;
}

bool check(ll n,ll pos) {return (n&(1<<pos));}
ll SET(ll n,ll pos){return (n|(1<<pos));}
ll RESET(ll n,ll pos) {return(n&~(1<<pos));}
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());

#define MOD 1000000007
#define MAX 1000005

ll h,w,ad[705][705];
string s[25];

int main()
{
    Fast
    FLR(i,700) FLR(j,700) ad[i][j]=INF;
    cin>>h>>w;
    map<pll,ll>mp;
    ll c = 1;
    FLR(i,h) cin>>s[i];
    FLR(i,h) FLR(j,w){
        if(s[i][j]=='.'){
            if(!mp[{i,j}]) mp[{i,j}]=c++;
            ll u = mp[{i,j}];
            if(i+1<h && s[i+1][j]=='.'){
                if(!mp[{i+1,j}]) mp[{i+1,j}]=c++;
                ll v = mp[{i+1,j}];
                ad[u][v]=1;
            }
            if(j+1<w && s[i][j+1]=='.'){
                if(!mp[{i,j+1}]) mp[{i,j+1}]=c++;
                ll v = mp[{i,j+1}];
                ad[u][v]=1;
            }
            if(i-1>=0 && s[i-1][j]=='.'){
                if(!mp[{i-1,j}]) mp[{i-1,j}]=c++;
                ll v = mp[{i-1,j}];
                ad[u][v]=1;
            }
            if(j-1>=0 && s[i][j-1]=='.'){
                if(!mp[{i,j-1}]) mp[{i,j-1}]=c++;
                ll v = mp[{i,j-1}];
                ad[u][v]=1;
            }
        }
    }
    repl(k,1,c) repl(i,1,c) repl(j,1,c) ad[i][j]=min(ad[i][j],ad[i][k]+ad[k][j]);
    ll mx = 0;
    repl(i,1,c) repl(j,1,c) mx = max(mx,(ad[i][j]==INF)?0:ad[i][j]);
    cout << mx << "\n";
}
/*
codingtine
*/