#include <bits/stdc++.h>
using namespace std;
const double PI = 4 * atan(1);
int N;
vector<int> adj[500013];
long long dp[500013];
long long ans;
void dfs(int n, int par = -1) {
  dp[n] = 1;
  for (int v : adj[n])
    if (v != par) {
      dfs(v, n);
      dp[n] = (998244353 + 1 - dp[v] * ((998244353 + 1) / 2) % 998244353) *
              dp[n] % 998244353;
    }
  ans = (ans + 998244353 + 1 - dp[n]) % 998244353;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N - 1; i++) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(0);
  for (int i = 0; i < N; i++) ans = (ans * 2) % 998244353;
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
const double PI = 4 * atan(1);
int N;
vector<int> adj[500013];
long long dp[500013];
long long ans;
void dfs(int n, int par = -1) {
  dp[n] = 1;
  for (int v : adj[n])
    if (v != par) {
      dfs(v, n);
      dp[n] = (998244353 + 1 - dp[v] * ((998244353 + 1) / 2) % 998244353) *
              dp[n] % 998244353;
    }
  ans = (ans + 998244353 + 1 - dp[n]) % 998244353;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N - 1; i++) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(0);
  for (int i = 0; i < N; i++) ans = (ans * 2) % 998244353;
  cout << ans << endl;
  return 0;
}

```