Problem:
Score : 400 points

Problem Statement

There is a grid with H horizontal rows and W vertical columns, and there are obstacles on some of the squares.

Snuke is going to choose one of the squares not occupied by an obstacle and place a lamp on it.
The lamp placed on the square will emit straight beams of light in four cardinal directions: up, down, left, and right.
In each direction, the beam will continue traveling until it hits a square occupied by an obstacle or it hits the border of the grid. It will light all the squares on the way, including the square on which the lamp is placed, but not the square occupied by an obstacle.

Snuke wants to maximize the number of squares lighted by the lamp.

You are given H strings S_i (1 \leq i \leq H), each of length W. If the j-th character (1 \leq j \leq W) of S_i is #, there is an obstacle on the square at the i-th row from the top and the j-th column from the left; if that character is ., there is no obstacle on that square.

Find the maximum possible number of squares lighted by the lamp.

Constraints

1 \leq H \leq 2,000

1 \leq W \leq 2,000

S_i is a string of length W consisting of # and ..

. occurs at least once in one of the strings S_i (1 \leq i \leq H).

Input

Input is given from Standard Input in the following format:

H W
S_1
:
S_H

Output

Print the maximum possible number of squares lighted by the lamp.

Sample Input 1

4 6
#..#..
.....#
....#.
#.#...

Sample Output 1

8

If Snuke places the lamp on the square at the second row from the top and the second column from the left, it will light the following squares: the first through fifth squares from the left in the second row, and the first through fourth squares from the top in the second column, for a total of eight squares.

Sample Input 2

8 8
..#...#.
....#...
##......
..###..#
...#..#.
##....#.
#...#...
###.#..#

Sample Output 2

13

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

#include <bits/stdc++.h>
//----***やべーやつら***----
using namespace std;
#define int long long

//----***型定義***----
using ll = long long;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
typedef long double lb;
// typedef vector<int,vector<int>> vvec;
typedef long double ld;
typedef pair<int,int> P;

//----***Like a Pythonista***----
#define REP(ii,jj,nn) for (ll ii=jj;ii<(nn);ii++)
#define RREP(ii,nn,jj) for (ll ii = nn; jj<ii;ii--)
#define each(i,...) for (auto&& i:__VA_ARGS__)
#define ALL(vec) (vec).begin(),(vec).end()
#define sum(...) accumulate(ALL(__VA_ARGS__),0LL)
#define dsum(...) accumulate(ALL(__VA_ARGS__),0.0L)
#define vec(type,name,...) vector<type> name(__VA_ARGS__)
template<class T> inline auto max(const T& a){ return *max_element(ALL(a)); }
template<class T> inline auto min(const T& a){ return *min_element(ALL(a)); }
inline ll gcd(ll a, ll b){ while(b){ ll c = b; b = a % b; a = c; } return a; }
inline ll lcm(ll a, ll b){ if(!a || !b) return 0; return a * b / gcd(a, b); }

//----***定数***----
const int MOD=1e9+7;
#define INF 1e9;
#define EPS 1e-9;

//----***入出力***---
#define print(out) cout<< out  << "\n";
#define debug(var)  do{std::cerr << #var << " ↓ "<<"\n";view(var);}while(0)
template<typename T> void view(T e){std::cout << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cout << e << " "; } std::cout << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){ for(const auto& v : vv){ view(v); } }
#define here  cout<<"Here"<<endl;


//----***初期時読み込み***----
struct initial{initial(){cin.tie(0); ios::sync_with_stdio(0); cout<<fixed<<setprecision(20);};}initial_;

signed main(){
  int H,W;cin>>H>>W;
  vector<string> graph(H);
  REP(i,0,H){
    cin>>graph[i];
  }
  vector<vector<int> > L(H,vector<int>(W,0)),R(H,vector<int>(W)),U(H,vector<int>(W)),D(H,vector<int>(W));
  REP(i,0,H)REP(j,0,W){
    if(graph[i][j]=='#'){L[i][j]=0;}
    else if(j==0){L[i][j]=1;}
    else {L[i][j]=L[i][j-1]+1;}
  }
  REP(i,0,H)RREP(j,W-1,-1){
    if (graph[i][j]=='#') {R[i][j]=0;}
    else if (j==W-1) {R[i][j]=1;}
    else {R[i][j]=R[i][j+1]+1;}
  }
  REP(i,0,H)REP(j,0,W){
    if(graph[i][j]=='#'){U[i][j]=0;}
    else if(i==0){U[i][j]=1;}
    else {U[i][j]=U[i-1][j]+1;}
  }
  RREP(i,H-1,-1)REP(j,0,W){
    if(graph[i][j]=='#'){D[i][j]=0;}
    else if(i==H-1){D[i][j]=1;}
    else {D[i][j]=D[i+1][j]+1;}
  }
  int ans=0;
  REP(i,0,H)REP(j,0,W){
    ans=max(ans,L[i][j]+R[i][j]+U[i][j]+D[i][j]-3);
  }
  print(ans)


}