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 <fstream>
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <random>
#include <numeric>

using namespace std;

#define INT_MAX_VALUE 2147483647
#define LONG_LONG_MAX_VALUE 9223372036854775807
#define ll long long
#define ld long double

struct XX{
    int b;
    ll c;
};

class xxGreater {
public:
    bool operator()(const XX& riLeft, const XX& riRight) const {
        //第2条件
        if((riLeft.c) == (riRight.c)){
            return riLeft.b > riRight.b;//<:昇順(小さいものから順番)、>:降順(大きいものから順番)
            //プライオリティキューの場合は > で、top()すると値の小さいものがとれる
        }
        //第1条件
        return (riLeft.c) > (riRight.c);
    }
};


//map<long long,long long> prime_f(long long n){
//    map<long long,long long>res;
//    for(int i=2;i*i<=n;i++){
//        while(n%i==0){
//            ++res[i];
//            n/=i;
//        }
//    }
//    if(n!=1)res[n]=1;
//    return res;
//}

//int n;
////int dat[2*10000000];
////int dat2[2*10000000];
//int dat[10];
//int dat2[10];
//
//void init(int n_){
//    n=1;
//    while(n<n_)n*=2;
//    for(int i=0;i<2*n-1;i++){
//        dat[i]=0;
//        dat2[i]=0;
//    }
//}
//
//void initset(int k,int a){
//    k+=n-1;
//    dat[k]=a;
//    while(k>0){
//        k=(k-1)/2;
//        dat[k]=dat[k*2+1]+dat[k*2+2];
//    }
//}
//
////[a,b)の間を[l,r]区間で比較しアップデート
////引数のindexに注意
////nは固定。initで計算すみ
////update2(L[i],R[i]+1,0,0,n,D[i]);
//void update2(int a,int b,int k,int l,int r,int v){//v更新値、区間は0-index
//    if(r<=a || b<=l)return;
//    if(a<=l && r<=b){
//        dat[k]+=dat2[k];
//        if(r-l>1){
//            dat2[k*2+1]+=dat2[k]/2;
//            dat2[k*2+1]+=dat2[k]/2;
//        }
//        dat2[k]=v*(r-l);
//        return;
//    }else{
//        update2(a,b,k*2+1,l,(l+r)/2,v);
//        update2(a,b,k*2+2,(l+r)/2,r,v);
//        return;
//    }
//}
//
//int query(int a,int b,int k,int l,int r){
//    if(r<=a || b<=l)return 0;
//    if(a<=l && r<=b){
//        dat[k]+=dat2[k];
//        if(r-l>1){
//            dat2[k*2+1]+=dat2[k]/2;
//            dat2[k*2+1]+=dat2[k]/2;
//        }
//        dat2[k]=0;
//        return dat[k];
//    }
//    else{
//        int vl=query(a,b,k*2+1,l,(l+r)/2);
//        int vr=query(a,b,k*2+2,(l+r)/2,r);
//        return vl+vr;
//    }
//}


int yoko[2000][2000];
int tate[2000][2000];
int main(int argc, const char * argv[])
{
    //scanf("%s",S);
    //scanf("%d",&N);
    //scanf("%lld %lld",&target1,&target2);
    //sscanf(tmp.c_str(),"%dd%d%d",&time[i], &dice[i], &z[i]);
    //getline(cin, target);
    //ifstream ifs("a.txt");//テスト用
    //ifs >> a;
    //ここから
    
    //入力高速化
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int H,W;
    cin>>H>>W;
    string s[2000];
    for(int i=0;i<H;i++){
        cin >> s[i];
    }
    
    for(int i=0;i<H;i++){
        int cnt=0;
        int start=0;
        for(int j=0;j<W;j++){
            if(s[i][j]=='#'){
                yoko[i][j]=0;
                for(int k=start;k<j;k++){
                    yoko[i][k]=cnt;
                }
                cnt=0;
                start=j+1;
            }else{
                cnt++;
            }
        }
        for(int k=start;k<W;k++){
            yoko[i][k]=cnt;
        }
    }
    for(int i=0;i<W;i++){
        int cnt=0;
        int start=0;
        for(int j=0;j<H;j++){
            if(s[j][i]=='#'){
                tate[j][i]=0;
                for(int k=start;k<j;k++){
                    tate[k][i]=cnt;
                }
                cnt=0;
                start=j+1;
            }else{
                cnt++;
            }
        }
        for(int k=start;k<H;k++){
            tate[k][i]=cnt;
        }
    }
    int ans=0;
    for(int i=0;i<H;i++){
        for(int j=0;j<W;j++){
            ans=max(ans,tate[i][j]+yoko[i][j]-1);
        }
    }
    cout << ans << endl;
    
    //ここまで
    //cout << "ans" << endl;
    //printf("%.0f\n",ans);//小数点以下表示なし
    //printf("%.7f\n",p);
    
    return 0;
}