#include <bits/stdc++.h>
#include <atcoder/all>

using namespace std;
using namespace atcoder;

using ll = long long;
using mint = modint998244353;
// using mint = modint1000000007;

template <typename T> using vec = vector<T>;
template <typename T> using pr = pair<T, T>;
template <typename T> using mipq = priority_queue<T, vec<T>, greater<T>>;

#define overload4(_1, _2, _3, _4, name, ...) name
#define rep1(i, n) for (auto i = decay_t<decltype(n)>{}; (i) < (n); ++(i))
#define rep2(i, l, r) for (auto i = (l); (i) < (r); ++(i))
#define rep3(i, l, r, d) for (auto i = (l); (i) < (r); i += (d))
#define rep(...) overload4(__VA_ARGS__, rep3, rep2, rep1)(__VA_ARGS__)
#define rrep(i, r, l) for (int i = (r); i >= (l); --i)
template <class T> bool chmax(T &a, const T b) { return (a < b ? a = b, true : false); }
template <class T> bool chmin(T &a, const T b) { return (a > b ? a = b, true : false); }

constexpr int INF = 1 << 30;
constexpr ll LINF = 1LL << 60;

void solve();

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);
    int T = 1;
    // cin >> T;
    while (T--) solve();
}

// 速度でさっき困りかけたのでC++
// bitごとに確率を求めればいいんだね？

void solve() {
    int N, K;
    cin >> N >> K;
    vec cnt(N, array<int, 30>{});

    rep(_, K) {
        int l, s;
        cin >> l >> s;
        vec<int> b(l);
        rep(i, l) cin >> b[i], b[i]--;
        rep(x, 30) if (s >> x & 1) {
            for (auto i : b) {
                cnt[i][x]++;
            }
        }
    }

    mint ans = 0;
    rep(i, N) {
        mint p = 1;
        rep(x, 30) {
            if (cnt[i][x] > 0) ans += p;
            p *= 2;
        }
    }
    ans /= 2;
    cout << ans.val() << '\n';
}