#include <bits/stdc++.h>
using namespace std;
const int N = 500005, Mod = 998244353;
int n, tot, sz[N], Pw[N], dp[N];
vector<int> Adj[N];
void DFS(int v, int p) {
  sz[v] = 1;
  for (int &u : Adj[v])
    if (u != p) DFS(u, v), sz[v] += sz[u];
  int cc = 1, left = sz[v] - 1;
  for (int &u : Adj[v])
    if (u != p) {
      left -= sz[u];
      dp[v] =
          (dp[v] + 1ll * Pw[left] * (Pw[sz[u] - 1] - dp[u] + Mod) % Mod * cc) %
          Mod;
      cc = 1ll * cc * (Pw[sz[u] - 1] + dp[u]) % Mod;
    }
  tot = (tot + 1ll * dp[v] * Pw[n - sz[v]]) % Mod;
}
int main() {
  scanf("%d", &n);
  for (int i = 1, v, u; i < n; i++)
    scanf("%d%d", &v, &u), Adj[v].push_back(u), Adj[u].push_back(v);
  for (int i = Pw[0] = 1; i < N; i++) Pw[i] = Pw[i - 1] * 2 % Mod;
  DFS(1, 0);
  return !printf("%d", (tot + tot) % Mod);
}

### 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;
const int N = 500005, Mod = 998244353;
int n, tot, sz[N], Pw[N], dp[N];
vector<int> Adj[N];
void DFS(int v, int p) {
  sz[v] = 1;
  for (int &u : Adj[v])
    if (u != p) DFS(u, v), sz[v] += sz[u];
  int cc = 1, left = sz[v] - 1;
  for (int &u : Adj[v])
    if (u != p) {
      left -= sz[u];
      dp[v] =
          (dp[v] + 1ll * Pw[left] * (Pw[sz[u] - 1] - dp[u] + Mod) % Mod * cc) %
          Mod;
      cc = 1ll * cc * (Pw[sz[u] - 1] + dp[u]) % Mod;
    }
  tot = (tot + 1ll * dp[v] * Pw[n - sz[v]]) % Mod;
}
int main() {
  scanf("%d", &n);
  for (int i = 1, v, u; i < n; i++)
    scanf("%d%d", &v, &u), Adj[v].push_back(u), Adj[u].push_back(v);
  for (int i = Pw[0] = 1; i < N; i++) Pw[i] = Pw[i - 1] * 2 % Mod;
  DFS(1, 0);
  return !printf("%d", (tot + tot) % Mod);
}

```