// 直径 * N TLE
#include <atcoder/modint.hpp>
#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
#define rep3(i, a, b, c) for (ll i = (a); i < (b); i += (c))
#define rep2(i, a, b) rep3(i, a, b, 1)
#define rep1(i, n) rep2(i, 0, n)
#define rep0(n) rep1(aaaaa, n)
#define ov4(a, b, c, d, name, ...) name
#define rep(...) ov4(__VA_ARGS__, rep3, rep2, rep1, rep0)(__VA_ARGS__)
#define per(i, a, b) for (ll i = (a) - 1; i >= (b); i--)
#define fore(e, v) for (auto &&e : v)
#define all(a) begin(a), end(a)
#define sz(a) (int)(size(a))

template <typename T, typename S> bool chmax(T &a, const S &b) {
  return a < b ? a = b, 1 : 0;
}
struct _ {
  _() { cin.tie(0)->sync_with_stdio(0), cout.tie(0); }
} __;

using ai3 = array<int, 3>;
using ai5 = array<int, 5>;
using mint = atcoder::modint998244353;

// a<b ?
bool greater_check(const ai3 &a, const ai3 &b) {
  ai5 abit, bbit;
  int ak = 0, bk = 0;
  rep(i, 3) ak += a[i];
  rep(i, 3) bk += b[i];
  rep(i, 3) abit[i] = a[i] + 1 + bk, bbit[i] = b[i] + 1 + ak;
  abit[3] = ak + 1, abit[4] = ak;
  bbit[3] = bk + 1, bbit[4] = bk;
  auto bitsort = [](ai5 &bit) {
    rep(i, 5) {
      rep(j, 1, 5) {
        if (bit[j] != -1 && bit[j] == bit[j - 1]) {
          bit[j - 1] = bit[j] + 1;
          bit[j] = -1;
        } else if (bit[j] > bit[j - 1]) {
          swap(bit[j], bit[j - 1]);
        }
      }
    }
  };

  bitsort(abit), bitsort(bbit);
  rep(i, 5) {
    if (abit[i] > bbit[i])
      return true;
    if (abit[i] < bbit[i])
      return false;
  }
  return false;
}

int main() {
  int N;
  cin >> N;
  vector<vi> G(N);
  rep(N - 1) {
    int U, V;
    cin >> U >> V;
    --U, --V;
    G[U].push_back(V);
    G[V].push_back(U);
  }
  auto bfs = [&](vi &look, vi &par, const vi &start) {
    fill(all(par), -1);
    look.clear();
    look.reserve(N);
    for (int i : start) {
      par[i] = i;
      look.push_back(i);
    }
    int l = 0;
    while (l < N) {
      int v = look[l++];
      for (int u : G[v]) {
        if (par[u] == -1) {
          par[u] = v;
          look.push_back(u);
        }
      }
    }
  };
  vi dist, par(N);
  bfs(dist, par, vi(1, 0));
  int d1 = dist.back();
  bfs(dist, par, vi(1, d1));
  int d2 = dist.back();
  vi diameter;
  diameter.reserve(N);
  int t = d2;
  do {
    diameter.push_back(t);
  } while (t == par[t] ? 0 : (t = par[t], 1));
  vi depth(N);
  ai3 ansd{0, 0, 0};
  t = 0;
  for (int i : diameter) {
    bfs(dist, par, vi(1, i));
    reverse(all(dist));
    fill(all(depth), 0);
    for (int i : dist) {
      if (par[i] != i)
        chmax(depth[par[i]], depth[i] + 1);
      else
        break;
    }
    vi nowd(2);
    for (auto d : G[i]) {
      nowd.push_back(depth[d] + 1);
    }
    sort(all(nowd), greater<int>());
    ai3 now{nowd[0], nowd[1], nowd[2]};
    if (greater_check(ansd, now))
      ansd = now;
    t++;
  }
  mint ans = mint(2).pow(N + 2);
  int K = 0;
  rep(i, 3) K += ansd[i];
  ans -= mint(2).pow(N - K - 1) *
         (mint(2).pow(ansd[0] + 2) + mint(2).pow(ansd[1] + 2) +
          mint(2).pow(ansd[2] + 2) - 6);
  cout << ans.val() << '\n';
}