#include <bits/stdc++.h>
using namespace std;
int n, f[505000], ans, iv2 = 998244353 + 1 >> 1;
vector<int> G[505000];
void dfs(int x, int fff) {
  f[x] = 1;
  for (auto y : G[x])
    if (y != fff) {
      dfs(y, x);
      f[x] = (1 - (long long)f[y] * iv2) % 998244353 * f[x] % 998244353;
    }
  ans = (ans + 1 - f[x]) % 998244353;
}
int main() {
  cin >> n;
  for (int i = n - 1, u, v; i--;) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) ans = ans * 2 % 998244353;
  cout << (ans + 998244353) % 998244353 << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int n, f[505000], ans, iv2 = 998244353 + 1 >> 1;
vector<int> G[505000];
void dfs(int x, int fff) {
  f[x] = 1;
  for (auto y : G[x])
    if (y != fff) {
      dfs(y, x);
      f[x] = (1 - (long long)f[y] * iv2) % 998244353 * f[x] % 998244353;
    }
  ans = (ans + 1 - f[x]) % 998244353;
}
int main() {
  cin >> n;
  for (int i = n - 1, u, v; i--;) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) ans = ans * 2 % 998244353;
  cout << (ans + 998244353) % 998244353 << endl;
}

```