Problem:
Score: 400 points

Problem statement

We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j).

Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares.

Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game.

When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares.

The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is .; if square (i, j) is initially painted by black, s_{i, j} is #.

Constraints

H is an integer between 2 and 50 (inclusive).

W is an integer between 2 and 50 (inclusive).

s_{i, j} is . or # (1 \leq i \leq H, 1 \leq j \leq W).

s_{1, 1} and s_{H, W} are ..

Input

Input is given from Standard Input in the following format:

H W
s_{1, 1}s_{1, 2}s_{1, 3} ... s_{1, W}
s_{2, 1}s_{2, 2}s_{2, 3} ... s_{2, W}
: :
s_{H, 1}s_{H, 2}s_{H, 3} ... s_{H, W}

Output

Print the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed.

Sample Input 1

3 3
..#
#..
...

Sample Output 1

2

The score 2 can be achieved by changing the color of squares as follows:

Sample Input 2

10 37
.....................................
...#...####...####..###...###...###..
..#.#..#...#.##....#...#.#...#.#...#.
..#.#..#...#.#.....#...#.#...#.#...#.
.#...#.#..##.#.....#...#.#.###.#.###.
.#####.####..#.....#...#..##....##...
.#...#.#...#.#.....#...#.#...#.#...#.
.#...#.#...#.##....#...#.#...#.#...#.
.#...#.####...####..###...###...###..
.....................................

Sample Output 2

209

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

#include<bits/stdc++.h>
#define all(x) (x).begin(),(x).end()
typedef long long ll;
#define rep(i,n) for(ll i=0, i##_len=(n); i<i##_len; ++i)
#define REP(i,num,n) for(ll i=num, i##_len=(n); i<i##_len; ++i)
#define repprev(i,a,b) for(ll i=b-1;i>=a;i--)
#define reprev(i,n) repprev(i,0,n)
using namespace std;
#define sz(x) ((int)(x).size())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define MEMSET(v, h) memset((v), h, sizeof(v))
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<class T> int former(const vector<T> &v, T x){
   return upper_bound(v.begin(),v.end(),x) - v.begin() - 1;
}
template<class T> int latter(const vector<T> &v, T x){
    return lower_bound(v.begin(),v.end(),x) - v.begin();
}
#define pb push_back
#define mp make_pair
#define y0 y3487465
#define y1 y8687969
#define j0 j1347829
#define j1 j234892
#define BIT_FLAG_0 (1<<0) // 0000 0000 0000 0001
#define BIT_FLAG_1 (1<<1) // 0000 0000 0000 0010
#define BIT_FLAG_2 (1<<2) // 0000 0000 0000 0100
#define BIT_FLAG_3 (1<<3) // 0000 0000 0000 1000
#define BIT_FLAG_4 (1<<4) // 0000 0000 0001 0000
#define BIT_FLAG_5 (1<<5) // 0000 0000 0010 0000
#define BIT_FLAG_6 (1<<6) // 0000 0000 0100 0000
#define BIT_FLAG_7 (1<<7) // 0000 0000 1000 0000
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
const ll LLINF = 1LL<<60;
const int INTINF = 1<<29;

const int MAX = 510000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

struct UnionFind {
    vector<ll> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(ll n) : par(n, -1) { }
    void init(ll n) { par.assign(n, -1); }

    ll root(ll x) {
        if (par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }

    bool issame(ll x, ll y) {
        return root(x) == root(y);
    }

    bool merge(ll x, ll y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (par[x] > par[y]) swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }

    ll size(ll x) {
        return -par[root(x)];
    }
};

template <typename T>
vector<T> dijkstra(int s,vector<vector<pair<int, T> > > & G){
   const T INF = numeric_limits<T>::max();
   using P = pair<T, int>;
   int n=G.size();
   vector<T> d(n,INF);
   vector<int> b(n,-1);
   priority_queue<P,vector<P>,greater<P> > q;
   d[s]=0;
   q.emplace(d[s],s);
   while(!q.empty()){
     P p=q.top();q.pop();
     int v=p.second;
     if(d[v]<p.first) continue;
     for(auto& e:G[v]){
       int u=e.first;
       T c=e.second;
       if(d[u]>d[v]+c){
         d[u]=d[v]+c;
         b[u]=v;
         q.emplace(d[u],u);
       }
     }
   }
   return d;
}

const int dx[4] = {1, 0, -1, 0}; // const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[4] = {0, 1, 0, -1}; // const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};


int main(){
  	int H,W;
  	cin >> H >> W;
  	int ans = H * W - H + (W-1);
  	char maze[51][51];
  	rep(i,H)
      rep(j,W)
      	cin >> maze[i][j];
  	
  	rep(i,H)
      rep(j,W)
      	if(maze[i][j] == '#') --ans;
  
  	cout << ans << endl;
}