#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e5 + 50;
vector<int> adj[MAXN];
int n;
void readInput() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
}
long long P[MAXN], E[MAXN], edge[MAXN], pow2[MAXN];
void findPow() {
  pow2[0] = 1;
  for (int k = 1; k <= n; k++) pow2[k] = (pow2[k - 1] * 2LL) % 998244353;
}
void dfs(int v, int p) {
  for (auto c : adj[v])
    if (c != p) dfs(c, v);
  for (auto c : adj[v])
    if (c != p) edge[v] += (edge[c] + 1);
  long long prod = 1;
  for (auto c : adj[v])
    if (c != p) prod = (prod * (pow2[edge[c]] + P[c])) % 998244353;
  P[v] = (pow2[edge[v]] - prod + 998244353) % 998244353;
  for (auto c : adj[v])
    if (c != p) E[v] = (E[v] + E[c]) % 998244353;
  E[v] = (E[v] + pow2[n - 1 - edge[v]] * P[v]) % 998244353;
}
int main() {
  readInput();
  findPow();
  dfs(1, -1);
  printf("%lld\n", (2 * E[1]) % 998244353);
}

### Prompt

Please provide a cpp coded solution 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 int MAXN = 5e5 + 50;
vector<int> adj[MAXN];
int n;
void readInput() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
}
long long P[MAXN], E[MAXN], edge[MAXN], pow2[MAXN];
void findPow() {
  pow2[0] = 1;
  for (int k = 1; k <= n; k++) pow2[k] = (pow2[k - 1] * 2LL) % 998244353;
}
void dfs(int v, int p) {
  for (auto c : adj[v])
    if (c != p) dfs(c, v);
  for (auto c : adj[v])
    if (c != p) edge[v] += (edge[c] + 1);
  long long prod = 1;
  for (auto c : adj[v])
    if (c != p) prod = (prod * (pow2[edge[c]] + P[c])) % 998244353;
  P[v] = (pow2[edge[v]] - prod + 998244353) % 998244353;
  for (auto c : adj[v])
    if (c != p) E[v] = (E[v] + E[c]) % 998244353;
  E[v] = (E[v] + pow2[n - 1 - edge[v]] * P[v]) % 998244353;
}
int main() {
  readInput();
  findPow();
  dfs(1, -1);
  printf("%lld\n", (2 * E[1]) % 998244353);
}

```