#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <time.h>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <cstring>

using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <iostream>

using namespace __gnu_pbds;
using namespace std;
template <class T>
using Tree =
    tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// to erase in multiset-> less_equal<T> and 
// s.erase(s.find_by_order(s.order_of_key(x)))
// lower_bound(x)=>(cannot use the stl lower_bound function)
// ll idx = s.order_of_key(x)
// if(idx == s.size()) -> no lower_bound
// else lb = *s.find_by_order(idx) // as 0-indexing
// idx-1 will give highest value which is strictly less than x
// for upper_bound->do the same with (x+1)

typedef long long ll;
typedef long double ld;
typedef pair<int,int> p32;
typedef pair<ll,ll> p64;
typedef tuple<ll, ll, ll> t64;
typedef vector<t64> vt64;
typedef vector<vt64> vvt64;
typedef pair<double,double> pdd;
typedef vector<ll> v64;
typedef vector<int> v32;
typedef vector<vector<int> > vv32;
typedef vector<vector<ll> > vv64;
typedef vector<vector<p64> > vvp64;
typedef vector<p64> vp64;
typedef vector<p32> vp32;
typedef vector<vector<p32> > vvp32;
typedef vector<bool> vb;
ll mod =  1e9+7, MOD = 998244353;
double eps = 1e-12;
#define FOR(s, e, i) for(ll i = s; i <= e; i++)
#define ROF(s ,e, i) for(ll i = s; i >= e; i--)
#define F0R(i, e) for(ll i = 0; i < (e); i++)
#define trav(e, a) for(auto &e : a)
#define coutAll(A) for(auto asdafas : A) cout <<  asdafas << " "; cout << "\n";
#define foutAll(A) for(auto asdafas : A) fout <<  asdafas << " "; cout << "\n";
#define cinAll(A) for(auto &asdafas : A) cin >>  asdafas;
#define finAll(A) for(auto &asdafas : A) fin >>  asdafas;
#define minpq priority_queue<ll, v64, greater<ll>>
#define maxpq priority_queue<ll> 
#define ln "\n"
#define dbg(x) cout<<#x<<" = "<<x<<ln
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define fi first
#define se second
ll inf = LLONG_MAX;
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())
#define yes cout<<"yes\n"
#define no cout<<"no\n"
#define Yes cout<<"Yes\n"
#define No cout<<"No\n"
#define YES cout<<"YES\n"
#define NO cout<<"NO\n"
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef pair<ll, ll> pll;
typedef pair<ll, ll> pii;
#define MAXN 100000000
ll powMod(ll x, ll y){
    ll res = 1;
    x %= MOD;
    while(y > 0){
        if(y&1) (res *= x) %= MOD;
        y >>= 1;
        (x *= x) %= MOD;
    }
    return res;
}
void solve(int it)
{
    ll n;
    cin >> n;
    vv64 adj(n);
    F0R(i, n - 1){
        ll u, v;
        cin >> u >> v;
        adj[--u].pb(--v);
        adj[v].pb(u);
    }
    v64 sub(n, 1), sub_sq_sum(n);
    auto dfs = [&](auto&&self, ll u, ll p = -1)->void{
        trav(v, adj[u]){
            if(v == p) continue;
            self(self, v, u);
            (sub[u] += sub[v]) %= MOD;
            (sub_sq_sum[u] += sub_sq_sum[v]) %= MOD;
        }
        (sub_sq_sum[u] += sub[u] * sub[u] % MOD) %= MOD;
    };
    dfs(dfs, 0);
    coutAll(sub_sq_sum);
    coutAll(sub);
    ll ans = 0;
    auto dfs2 = [&](auto&&self, ll u, ll p = -1)->void{
        ll a = 0, b = 0;
        trav(v, adj[u]){
            if(v == p) continue;
            // keep u-v
            // finding the ei^2 contribution
            (ans += sub[v]%MOD*sub[v]%MOD*(n-sub[v])%MOD*(n-sub[v])%MOD) %= MOD;
            // finding ei*ej contribution
            // with one edge with v, another in the subtree of v
            ans += 2*(sub_sq_sum[v]-sub[v]*sub[v]%MOD+MOD)%MOD*(n-sub[v])%MOD*(n-sub[v])%MOD;
            ans %= MOD;
            // both in different child's subtree
            a += sub_sq_sum[v]; a%=MOD;
            b += sub_sq_sum[v]*sub_sq_sum[v]%MOD; b %= MOD;
            self(self, v, u);
        }
        // cout << u+1 <<" " << a << " " << b << " " << a*a-b << "\n";
        ans += (a*a%MOD - b + MOD)%MOD%MOD;
        ans %= MOD;
    };
    dfs2(dfs2, 0);
    cout << ans << "\n";
    ans *= powMod(powMod(n*(n-1)/2, 2), MOD-2);
    ans %= MOD;
    cout << ans << "\n";    
}


int main()
{
    fast_cin();    
    ll t = 1;
    cin >> t;
    for(int it=1; it<=t; it++)
    {
        //cout << "Case " << it << ": ";
        solve(it);
    }
    return 0;
}