#include <bits/stdc++.h>
using namespace std;
const int P = 998244353, iv2 = 499122177;
int n, m;
int A[500006];
vector<int> G[500006];
int dp[500006], as = 0;
void dfs(int u, int f) {
  int p = 1;
  for (int v : G[u])
    if (v != f) {
      dfs(v, u);
      p = p * 1ll * (1 + dp[v]) % P * iv2 % P;
    }
  dp[u] = 1 + P - p;
  as = (as + dp[u]) % P;
}
void solve() {
  cin >> n;
  int p2 = 2;
  for (int i = (2), iend = (n); i <= iend; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v), G[v].push_back(u);
    p2 = p2 * 2 % P;
  }
  dfs(1, 1);
  cout << as * 1ll * p2 % P << endl;
}
signed main() { solve(); }

### Prompt

Create a solution in Cpp for 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 int P = 998244353, iv2 = 499122177;
int n, m;
int A[500006];
vector<int> G[500006];
int dp[500006], as = 0;
void dfs(int u, int f) {
  int p = 1;
  for (int v : G[u])
    if (v != f) {
      dfs(v, u);
      p = p * 1ll * (1 + dp[v]) % P * iv2 % P;
    }
  dp[u] = 1 + P - p;
  as = (as + dp[u]) % P;
}
void solve() {
  cin >> n;
  int p2 = 2;
  for (int i = (2), iend = (n); i <= iend; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v), G[v].push_back(u);
    p2 = p2 * 2 % P;
  }
  dfs(1, 1);
  cout << as * 1ll * p2 % P << endl;
}
signed main() { solve(); }

```