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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define SWAP(type, x, y) do { type tmp = x; x = y; y = tmp; } while (0)

#define MOD 1000000007 

typedef struct obj_{
    int key;
    struct obj_ *next;
}*obj;

//slobj型の変数のポインタをslist型と定義→いずれリストの先頭・最後尾要素を指す
typedef struct{
    obj head;
    obj tail;
}*queue;

//keyにxをもつキュー要素を作る関数
obj obj_new(int x){
    obj p;
    NEW(p,1);
    p->key=x;
    p->next=NULL;
    return p;
}

//slist型の空リスト1つを作る関数
queue queue_new(void){
    queue Q;
    NEW(Q,1);
    Q->head=NULL;
    Q->tail=NULL;
    return Q;
}

//enqueue(リスト最後尾に挿入)
void enqueue(queue Q,obj p){
    if(Q->head==NULL){
        Q->head=p;
        Q->tail=p;
    }
    else{
        Q->tail->next=p;
        Q->tail=p;
    }
    
    return;
}

obj dequeue(queue Q){
    obj p=Q->head;
    if(p==Q->tail){
        Q->tail=NULL;
        Q->head=NULL;
    }
    else{
        Q->head=p->next;
    }
    return p;
}

int main(void){
    int H,W;
    scanf("%d%d",&H,&W);

    int M[H][W];
    int white=0;
    for(int i=0;i<H;i++){
        char c[W+1];
        scanf("%s",c);
        for(int j=0;j<W;j++){
            if(c[j]=='.'){
                M[i][j]=0;
                white++;
            }
            else M[i][j]=-1; 
        }
    }

    queue Q=queue_new();
    enqueue(Q,obj_new(0));
    while(1){
        obj p=dequeue(Q);
        if(p==NULL) break;

        //これからM[x][y]の話をする(対応はx*W+y)
        int x=(p->key)/W;
        int y=(p->key)%W;
        if(x!=0 && M[x-1][y]==0){
            enqueue(Q,obj_new((x-1)*W+y));
            M[x-1][y]=M[x][y]+1;
        }
        if(x!=H-1 && M[x+1][y]==0){
            enqueue(Q,obj_new((x+1)*W+y));
            M[x+1][y]=M[x][y]+1;
        }
        if(y!=0 && M[x][y-1]==0){
            enqueue(Q,obj_new(x*W+y-1));
            M[x][y-1]=M[x][y]+1;
        }
        if(y!=W-1 && M[x][y+1]==0){
            enqueue(Q,obj_new(x*W+y+1));
            M[x][y+1]=M[x][y]+1;
        }
    }

    int ans=white-(M[H-1][W-1]+1);
    printf("%d\n",ans);

    return 0;
}