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 PyPy3 (2.4.0) (judge verdict: Accepted)

from sys import stdin
from collections import deque
from collections import Counter


def grid2adj(arr_grid,direction=4):
    def cand(i,j,direction):
        if direction == 4:
            return [[i-1,j],[i+1,j],[i,j-1],[i,j+1]]
        elif direction == 8:
            return [[i-1,j],[i+1,j],[i,j-1],[i,j+1],
                    [i-1,j-1],[i-1,j+1],[i+1,j-1],[i+1,j+1]]
        else:
            print('Error')
    
    M,N = len(arr_grid),len(arr_grid[0])
    arr_con = [[] for i in range(M*N)]
    for i in range(M):
        for j in range(N):
            v = i*N+j
            for _cand in cand(i,j,direction):
                if 0<=_cand[0]<M and 0<=_cand[1]<N:
                    next_v = _cand[0]*N+_cand[1]
                    arr_con[v].append(next_v)
    arr_val = sum(arr_grid,[])

    return arr_con, arr_val


class BFS():
    def __init__(self, con, val=None):
        # 隣接リスト表現
        self.con = con
        self.val = val
        self.N = len(self.con)
        # 始点からの距離
        self.dist = [-1]*self.N
        self.que = deque()

    def bfs(self, start_v):
        # 初期条件(0-indexed)
        self.dist[start_v] = 0
        self.que.append(start_v)

        while len(self.que)>0:
            v = self.que.popleft()
            for next_v in self.con[v]:
                if self.dist[next_v] != -1:
                    continue
                if self.val[next_v] == '#':
                    continue
                self.dist[next_v] = self.dist[v] + 1
                self.que.append(next_v)


if __name__ == "__main__":
    _in = [_.rstrip() for _ in stdin.readlines()]
    H, W = list(map(int,_in[0].split(' ')))  # type:list(int)
    graph = []
    for i in range(H):
        _ = list(_in[i+1])  # type:list(str)
        graph.append(_)
    # vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
    g_con, g_val = grid2adj(graph)
    bfs = BFS(g_con,g_val)

    cnt = 0
    bfs.bfs(0)
    if bfs.dist[-1] == -1:
        cnt = -1
    else:
        cnt = H*W-bfs.val.count('#')-(bfs.dist[-1]+1)
    # ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    print(cnt)