using namespace std;

int main() {
    ios::sync_with_stdio(false);
        cin.tie(nullptr);
            freopen("hiking.in", "r", stdin);
                freopen("hiking.out", "w", stdout);

                    int n, m;
                        cin >> n >> m;
                            vector<string> g(n);
                                for (int i = 0; i < n; ++i) cin >> g[i];

                                    vector<vector<int>> vis(n, vector<int>(m));
                                        int ans = 0;
                                            int dx[4] = {-1, 1, 0, 0}, dy[4] = {0, 0, -1, 1};

                                                for (int i = 0; i < n; ++i) {
                                                        for (int j = 0; j < m; ++j) {
                                                                    if (g[i][j] != '#' || vis[i][j]) continue;
                                                                                ++ans;
                                                                                            queue<pair<int,int>> q;
                                                                                                        q.push({i, j});
                                                                                                                    vis[i][j] = 1;
                                                                                                                                while (!q.empty()) {
                                                                                                                                                auto [x, y] = q.front(); q.pop();
                                                                                                                                                                for (int k = 0; k < 4; ++k) {
                                                                                                                                                                                    int nx = x + dx[k], ny = y + dy[k];
                                                                                                                                                                                                        if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;
                                                                                                                                                                                                                            if (vis[nx][ny] || g[nx][ny] != '#') continue;
                                                                                                                                                                                                                                                vis[nx][ny] = 1;
                                                                                                                                                                                                                                                                    q.push({nx, ny});
                                                                                                                                                                                                                                                                                    }
                                                                                                                                                                                                                                                                                                }
                                                                                                                                                                                                                                                                                                        }
                                                                                                                                                                                                                                                                                                            }

                                                                                                                                                                                                                                                                                                                cout << ans << '\n';
                                                                                                                                                                                                                                                                                                                    return 0;
                                                                                                                                                                                                                                                                                                                    }****