#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
const int N = 5 * (int)1e5;
vector<int> conns[N];
long long res = 0;
long long dp[N];
long long modPow(long long a, long long b) {
  if (b == 0) return 1;
  if (b & 1) return (a * modPow(a, b - 1)) % MOD;
  return modPow(a * a % MOD, b / 2);
}
long long modInv(long long v) { return modPow(v, MOD - 2); }
void solve(int i, int p) {
  long long pc = 1;
  for (auto t : conns[i]) {
    if (t == p) continue;
    solve(t, i);
    pc = (pc * (1 - dp[t])) % MOD;
  }
  res = (res + (1 - pc)) % MOD;
  dp[i] = (modInv(2) * pc) % MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    conns[a].push_back(b), conns[b].push_back(a);
  }
  solve(0, 0);
  for (int i = 0; i < n; ++i) res = (res * 2) % MOD;
  if (res < 0) res += MOD;
  cout << res << '\n';
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
const int N = 5 * (int)1e5;
vector<int> conns[N];
long long res = 0;
long long dp[N];
long long modPow(long long a, long long b) {
  if (b == 0) return 1;
  if (b & 1) return (a * modPow(a, b - 1)) % MOD;
  return modPow(a * a % MOD, b / 2);
}
long long modInv(long long v) { return modPow(v, MOD - 2); }
void solve(int i, int p) {
  long long pc = 1;
  for (auto t : conns[i]) {
    if (t == p) continue;
    solve(t, i);
    pc = (pc * (1 - dp[t])) % MOD;
  }
  res = (res + (1 - pc)) % MOD;
  dp[i] = (modInv(2) * pc) % MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    conns[a].push_back(b), conns[b].push_back(a);
  }
  solve(0, 0);
  for (int i = 0; i < n; ++i) res = (res * 2) % MOD;
  if (res < 0) res += MOD;
  cout << res << '\n';
}

```