# 盤面が非常に小さい、visited管理してDFSにしよう
# 問題文では「没収されるときに1つも宝石を持っていない状況」とあるが0個になるのはOKなのか
# 動きが上下左右だけというのも書いてない

H, W = map(int, input().split())
A = []
for i in range(H):
    A.append(input())
    
import sys
sys.setrecursionlimit(10**7)

def dfs(ch, cw, step, coin):
    global ans_list, min_step    
    
    #print('ch', ch, 'cw', cw, 'step', step, 'coin', coin)
    #print('visited', visited)
    #print()
    if step > min_step:
        return

    if (ch, cw)==(H-1, W-1):
        ans_list.append(step)
        if step < min_step:
            min_step = step
        return
    
    for dh, dw in d:
        if 0 <= ch+dh < H and 0 <= cw+dw < W and visited[ch+dh][cw+dw ]==0:
            if A[ch+dh][cw+dw] == 'o':
                visited[ch+dh][cw+dw ]=1
                dfs(ch+dh, cw+dw, step+1, coin+1)
                visited[ch+dh][cw+dw ]=0
            elif A[ch+dh][cw+dw] == 'x':
                if coin>=1:
                    visited[ch+dh][cw+dw ]=1
                    dfs(ch+dh, cw+dw, step+1, coin-1)
                    visited[ch+dh][cw+dw ]=0
    
d = [[+1, 0], [-1, 0], [0, +1], [0, -1]]
min_step = 10**3
ans_list = []
visited = [[0]*W for i in range(H)]
visited[0][0] = 1
dfs(0, 0, 0, 1)
#print(ans_list)

if len(ans_list)>0:
    min_step = min(ans_list)
    ans = ans_list.count(min_step)
else:
    ans = 0
print(ans)