#include <bits/stdc++.h>
#ifdef LOCAL
#include <debug.hpp>
#else
#define debug(...) void(0)
#endif

#include "atcoder/modint"

using namespace std;

typedef long long ll;
#define all(x) begin(x), end(x)
constexpr int INF = (1 << 30) - 1;
constexpr long long IINF = (1LL << 60) - 1;
constexpr int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};

template <class T> istream& operator>>(istream& is, vector<T>& v) {
    for (auto& x : v) is >> x;
    return is;
}

template <class T> ostream& operator<<(ostream& os, const vector<T>& v) {
    auto sep = "";
    for (const auto& x : v) os << exchange(sep, " ") << x;
    return os;
}

template <class T, class U = T> bool chmin(T& x, U&& y) { return y < x and (x = forward<U>(y), true); }

template <class T, class U = T> bool chmax(T& x, U&& y) { return x < y and (x = forward<U>(y), true); }

template <class T> void mkuni(vector<T>& v) {
    sort(begin(v), end(v));
    v.erase(unique(begin(v), end(v)), end(v));
}

template <class T> int lwb(const vector<T>& v, const T& x) { return lower_bound(begin(v), end(v), x) - begin(v); }

using mint = atcoder::modint998244353;

mint f(int n) { return mint(n) * (n + 1) / 2; }

void solve() {
    int n;
    cin >> n;
    vector<vector<int>> G(n);
    for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        u--, v--;
        G[u].emplace_back(v);
        G[v].emplace_back(u);
    }

    mint ans = 0;
    vector<int> sub(n);
    vector<mint> sq(n), prob(n);  // v が白く残る確率
    mint tot = f(n);
    auto dfs = [&](auto self, int v, int p) -> void {
        sub[v] = 1;
        mint sum = 0;
        sq[v] = 0;
        for (int& u : G[v]) {
            if (u == p) continue;
            self(self, u, v);
            sub[v] += sub[u];
            sum += f(sub[u]);
            sq[v] += f(sub[u]);
        }
        sum += f(n - sub[v]);
        prob[v] = (tot - sum) / tot;
        debug(v, prob[v].val());
        ans += (1 - prob[v]).inv();
        for (int& u : G[v]) {
            if (u == p) continue;
            mint prob2 = 1;
            prob2 -= 1 - prob[v];
            prob2 -= 1 - prob[u];
            mint SUM = sum;
            SUM -= f(sub[u]);
            SUM += sq[u];
            SUM /= tot;
            prob2 += SUM;
            ans -= (1 - prob2).inv();
            debug(v, u, prob2.val());
        }
    };
    dfs(dfs, 0, -1);

    cout << ans.val() << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T;
    cin >> T;
    for (; T--;) solve();
    return 0;
}