# oが少ない順に処理する (誤解法)
from collections import Counter
N, M = map(int, input().split())
ans = 0
end = set()
S = [input() for _ in range(N)]
Count = [(c["o"], i) for i, c in enumerate([Counter(j) for j in S])]
Count.sort()
for i in range(M):
    C = []
    for j in Count:
        j = j[1]
        if S[j][i] == "o":
            C.append(j)
	
    if len(C) == 1 and C[0] not in end:
        end.add(C[0])
        ans += 1
    else:
        for c in C:
            end.add(c)

print(ans)