#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
void ckmin(T &a, U b) {
  if (a > b) a = b;
}
template <class T, class U>
void ckmax(T &a, U b) {
  if (a < b) a = b;
}
int N;
vector<int> edge[500013];
pair<long long, long long> dp[500013];
long long ans;
void dfs(int u, int p) {
  dp[u] = {0, 1};
  for (int v : edge[u]) {
    if (v == p) continue;
    dfs(v, u);
    dp[u].first += (dp[v].first + (998244353 + 1 - dp[v].second));
    dp[u].first %= 998244353;
    dp[u].second *=
        (499122177 + 499122177 * (998244353 + 1 - dp[v].second) % 998244353);
    dp[u].second %= 998244353;
  }
}
int32_t main() {
  cout << fixed << setprecision(12);
  cerr << fixed << setprecision(4);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (auto i = (0); i < (N - 1); i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  dfs(0, N);
  ans = dp[0].first + (998244353 + 1 - dp[0].second);
  ans %= 998244353;
  for (auto i = (0); i < (N); i++) {
    ans += ans;
    if (ans >= 998244353) ans -= 998244353;
  }
  cout << ans << '\n';
}

### Prompt

Your task is to create a Cpp solution to the following problem:

Let's define rank of undirected graph as rank of its adjacency matrix in R^{n × n}.

Given a tree. Each edge of this tree will be deleted with probability 1/2, all these deletions are independent. Let E be the expected rank of resulting forest. Find E ⋅ 2^{n-1} modulo 998244353 (it is easy to show that E ⋅ 2^{n-1} is an integer).

Input

First line of input contains n (1 ≤ n ≤ 5 ⋅ 10^{5}) — number of vertices.

Next n-1 lines contains two integers u v (1 ≤ u,    v ≤ n;    u ≠ v) — indices of vertices connected by edge.

It is guaranteed that given graph is a tree.

Output

Print one integer — answer to the problem.

Examples

Input

3
1 2
2 3


Output

6


Input

4
1 2
1 3
1 4


Output

14


Input

4
1 2
2 3
3 4


Output

18

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
void ckmin(T &a, U b) {
  if (a > b) a = b;
}
template <class T, class U>
void ckmax(T &a, U b) {
  if (a < b) a = b;
}
int N;
vector<int> edge[500013];
pair<long long, long long> dp[500013];
long long ans;
void dfs(int u, int p) {
  dp[u] = {0, 1};
  for (int v : edge[u]) {
    if (v == p) continue;
    dfs(v, u);
    dp[u].first += (dp[v].first + (998244353 + 1 - dp[v].second));
    dp[u].first %= 998244353;
    dp[u].second *=
        (499122177 + 499122177 * (998244353 + 1 - dp[v].second) % 998244353);
    dp[u].second %= 998244353;
  }
}
int32_t main() {
  cout << fixed << setprecision(12);
  cerr << fixed << setprecision(4);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (auto i = (0); i < (N - 1); i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  dfs(0, N);
  ans = dp[0].first + (998244353 + 1 - dp[0].second);
  ans %= 998244353;
  for (auto i = (0); i < (N); i++) {
    ans += ans;
    if (ans >= 998244353) ans -= 998244353;
  }
  cout << ans << '\n';
}

```