#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353, inv2 = MOD / 2 + 1;
const int maxn = 5e5 + 9;
int N;
long long dp[maxn], ans = 0, pw2_N = 2;
vector<int> g[maxn];
void dfs(int x, int fa = -1) {
  dp[x] = 1;
  for (auto e : g[x]) {
    if (e == fa) continue;
    dfs(e, x);
    dp[x] = (1 - inv2 * dp[e] % MOD + MOD) * dp[x] % MOD;
  }
  ans += pw2_N * (MOD + 1 - dp[x]) % MOD;
}
int main() {
  cin >> N;
  for (int i = 1, u, v; i < N; i++) {
    pw2_N = pw2_N * 2 % MOD;
    scanf("%d%d", &u, &v);
    --u, --v;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  dfs(0);
  cout << ans % MOD << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
const long long MOD = 998244353, inv2 = MOD / 2 + 1;
const int maxn = 5e5 + 9;
int N;
long long dp[maxn], ans = 0, pw2_N = 2;
vector<int> g[maxn];
void dfs(int x, int fa = -1) {
  dp[x] = 1;
  for (auto e : g[x]) {
    if (e == fa) continue;
    dfs(e, x);
    dp[x] = (1 - inv2 * dp[e] % MOD + MOD) * dp[x] % MOD;
  }
  ans += pw2_N * (MOD + 1 - dp[x]) % MOD;
}
int main() {
  cin >> N;
  for (int i = 1, u, v; i < N; i++) {
    pw2_N = pw2_N * 2 % MOD;
    scanf("%d%d", &u, &v);
    --u, --v;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  dfs(0);
  cout << ans % MOD << endl;
  return 0;
}

```