#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <utility>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <tuple>
#include <numeric>
#include <functional>
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef pair<ll, ll> P;
#define rep(i, n) for(ll i = 0; i < n; i++)
#define exrep(i, a, b) for(ll i = a; i <= b; i++)
#define out(x) cout << x << endl
#define exout(x) printf("%.10f\n", x)
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define pb push_back
#define re0 return 0
const ll mod = 998244353;
const ll INF = 1e16;
 
const ll MAX = 100010;
 
// a^n (mod.MOD)を求める。計算量はO(logn)
ll modpow(ll a, ll n, ll MOD = mod) {
    if(n == 0) {
        return 1;
    }
    if(n%2 == 1) {
        return (a * modpow(a, n-1, MOD)) % MOD;
    }
    return (modpow(a, n/2, MOD) * modpow(a, n/2, MOD)) % MOD;
}
 
ll inverse(ll a) {
    return modpow(a, mod - 2);
}
 
ll fact[MAX];  // fact[i] : iの階乗のmod
ll inv[MAX];  // inv[i] : iの逆数のmod
ll invfact[MAX];  // invfact[i] : iの階乗の逆数のmod

void init() {
    fact[0] = 1;
    inv[0] = inv[1] = 1;
    invfact[0] = 1;
    for(ll i = 1; i < MAX; i++) {
        fact[i] = i * fact[i-1] % mod;
        if(i >= 2) {
            inv[i] = mod - inv[mod % i] * (mod / i) % mod;
        }
        invfact[i] = inverse(fact[i]);
    }
}
 
// nCrをO(n)で求める。 
ll Comb(ll n, ll r) {
    if(r < 0 || n < 0 || n < r) {
        return 0;
    }
    ll res = fact[n];
    res = (res * invfact[r]) % mod;
    res = (res * invfact[n-r]) % mod;
    return res;
}
 
ll n;
vvl G;
vl sz;  // sz[v] : vを根とする部分木のサイズ
vl vec;
 
void dfs1(ll v, ll p = -1) {
    for(ll u : G[v]) {
        if(u == p) { continue; }
        dfs1(u, v);
        sz[v] += sz[u];
    }
}
 
void dfs2(ll v, ll p = -1) {
    vec.pb(n - sz[v]);
    for(ll u : G[v]) {
        if(u == p) { continue; }
        vec.pb(sz[u]);
        dfs2(u, v);
    }
}
 
int main() {
    
    cin >> n;    
    
    G.resize(n);
    rep(i, n-1) {
        ll a, b;
        cin >> a >> b;
        a--;  b--;
        G[a].pb(b);
        G[b].pb(a);
    }
 
    sz.assign(n, 1);
    dfs1(0);
 
    dfs2(0);
 
    init();
    
    ll x = 0;
    for(ll y : vec) {
        x = (x + Comb(y, 2)) % mod;
    }

    ll ans = 2 * x * inv[n] % mod * inv[n-1] % mod * inv[n-1] % mod;
    
    out(ans);
    re0;
}