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<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=100;
bool vst[maxn][maxn]; // 访问标记
int dir[4][2]={0,1,0,-1,1,0,-1,0}; // 方向向量
int qaq,qmq,ans=0;
char qwq[maxn][maxn];
struct State // BFS 队列中的状态数据结构
{
int x,y; // 坐标位置
int Step_Counter; // 搜索步数统计器
};
bool CheckState(State s) // 约束条件检验
{
if(!vst[s.x][s.y] &&qwq[s.x][s.y]=='.') // 满足条件
return 1;
else // 约束条件冲突
return 0;
}

void bfs(State st)
{
queue <State> q; // BFS 队列
State now,next; // 定义2 个状态，当前和下一个
st.Step_Counter=0; // 计数器清零
q.push(st); // 入队
vst[st.x][st.y]=1; // 访问标记
while(!q.empty())
{
now=q.front(); // 取队首元素进行扩展
if(now.x==qaq&&now.y==qmq) // 出现目标态，此时为Step_Counter 的最小值，可以退出即可
{
	printf("%d",qaq*qmq-ans-1-now.Step_Counter);
 // 做相关处理
return;
}
for(int i=0;i<4;i++)
{
next.x=now.x+dir[i][0]; // 按照规则生成下一个状态
next.y=now.y+dir[i][1];
next.Step_Counter=now.Step_Counter+1; // 计数器加1
if(CheckState(next)) // 如果状态满足约束条件则入队
{
q.push(next);
vst[next.x][next.y]=1; //访问标记
}
}
q.pop(); // 队首元素出队
}
 return;
}

int main(){
	scanf("%d %d",&qaq,&qmq);
	for(int i=1;i<=qaq;i++)
		for(int j=1;j<=qmq;j++)
			scanf(" %c",&qwq[i][j]);
	for(int i=1;i<=qaq;i++)
		for(int j=1;j<=qmq;j++)
			if(qwq[i][j]=='#')ans++;
	State a;
	a.x=1;a.y=1;
	bfs(a);
return 0;
}