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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.StringTokenizer;

class Main {

    public static void main(String[] args) throws IOException {
        ReaderD.init(System.in);
        int h = ReaderD.nextInt();
        int w = ReaderD.nextInt();
        char[][] grid = new char[h][w];
        for (int i = 0; i < grid.length; i++) grid[i] = ReaderD.next().toCharArray();
        Graph gridGraph = new Graph(h*w);
        int white = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                if (grid[i][j] == '.') {
                    white++;
                    if (i + 1 < grid.length && grid[i + 1][j] == '.') gridGraph.addEdge(i*w + j, (i + 1)*w + j);
                    if (j + 1 < grid[i].length && grid[i][j + 1] == '.') gridGraph.addEdge(i*w + j, i*w + j + 1);
                }
            }
        }

        gridGraph.bfs(0);
        if (gridGraph.distTo[h*w - 1] == -1) System.out.println(-1);
        else System.out.println(white - gridGraph.distTo[h*w - 1] - 1);
    }
}

class Graph{
    private int V;
    private ArrayList<Integer> adj[];
    private boolean visitedBFS[];
    private LinkedList<Integer> queue;
    int[] distTo;

    Graph(int V){
        this.V = V;
        adj = (ArrayList<Integer>[]) new ArrayList[V];
        visitedBFS = new boolean[V];
        distTo = new int[V];
        Arrays.fill(distTo, -1);
        queue = new LinkedList<>();
        for (int i = 0; i < V; i++) adj[i] = new ArrayList<>();
    }

    int vertices(){
        return V;
    }

    void bfs(int x){
        visitedBFS[x] = true;
        queue.addLast(x);
        distTo[x] = 0;
        while (!queue.isEmpty()){
            int s = queue.getFirst(); queue.removeFirst();
            for (int u: adj[s]){
                if (visitedBFS[u]) continue;
                visitedBFS[u] = true;
                queue.addLast(u);
                distTo[u] = 1 + distTo[s];
            }
        }
    }

    void addEdge(int v, int w){
        adj[v].add(w);
        adj[w].add(v);
    }

    Iterable<Integer>[] getAdj(){
        return adj;
    }
}

/**
 * Reader class based on the article at "https://www.cpe.ku.ac.th/~jim/java-io.html"
 */
class ReaderD {
    private static BufferedReader reader;
    private static StringTokenizer tokenizer;

    static void init(InputStream inputStream) {
        reader = new BufferedReader(new InputStreamReader(inputStream));
        tokenizer = new StringTokenizer("");
    }

    static String next() throws IOException {
        String read;
        while (!tokenizer.hasMoreTokens()) {
            read = reader.readLine();
            if (read == null || read.equals(""))
                return "-1";
            tokenizer = new StringTokenizer(read);
        }

        return tokenizer.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    static long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    //    Get a whole line.
    static String line() throws IOException {
        return reader.readLine();
    }

    static double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
}