#include <bits/stdc++.h>
using namespace std;
long long dp[500005];
;
long long bin[500005];
int n;
vector<int> edge[500005];
int cnt[500005];
long long ans;
void dfs(int v, int u) {
  cnt[v] = 1;
  for (int i = 0; i < edge[v].size(); i++) {
    if (edge[v][i] == u) continue;
    dfs(edge[v][i], v);
    cnt[v] += cnt[edge[v][i]];
  }
  dp[v] = 1;
  int cur = 1;
  for (int i = 0; i < edge[v].size(); i++) {
    if (edge[v][i] == u) continue;
    int sz = cnt[edge[v][i]];
    cur += sz;
    ans += dp[v] * dp[edge[v][i]] % 998244353 * bin[n - cur] % 998244353;
    dp[v] *= bin[sz] - dp[edge[v][i]];
    dp[v] %= 998244353;
  }
}
int main() {
  bin[0] = 1;
  for (int i = 1; i < 500005; i++) bin[i] = bin[i - 1] * 2LL % 998244353;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  dfs(1, -1);
  cout << (2LL * ans % 998244353 + 998244353) % 998244353 << endl;
}

### 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;
long long dp[500005];
;
long long bin[500005];
int n;
vector<int> edge[500005];
int cnt[500005];
long long ans;
void dfs(int v, int u) {
  cnt[v] = 1;
  for (int i = 0; i < edge[v].size(); i++) {
    if (edge[v][i] == u) continue;
    dfs(edge[v][i], v);
    cnt[v] += cnt[edge[v][i]];
  }
  dp[v] = 1;
  int cur = 1;
  for (int i = 0; i < edge[v].size(); i++) {
    if (edge[v][i] == u) continue;
    int sz = cnt[edge[v][i]];
    cur += sz;
    ans += dp[v] * dp[edge[v][i]] % 998244353 * bin[n - cur] % 998244353;
    dp[v] *= bin[sz] - dp[edge[v][i]];
    dp[v] %= 998244353;
  }
}
int main() {
  bin[0] = 1;
  for (int i = 1; i < 500005; i++) bin[i] = bin[i - 1] * 2LL % 998244353;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  dfs(1, -1);
  cout << (2LL * ans % 998244353 + 998244353) % 998244353 << endl;
}

```