#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
struct expe {
  int sum, cnt;
  expe(long long s = 0, long long c = 0) { sum = s % mod, cnt = c % mod; }
  expe operator+(expe x) { return expe(sum + x.sum, cnt + x.cnt); }
  expe operator*(expe x) {
    return expe(1ll * sum * x.cnt + 1ll * x.sum * cnt, 1ll * cnt * x.cnt);
  }
} dp[500050][2];
int n;
vector<int> adj[500050];
void dfs(int u, int pa = 0) {
  dp[u][1] = expe(0, 1);
  for (auto v : adj[u]) {
    if (v == pa) continue;
    dfs(v, u);
    expe x = dp[u][0], y = dp[u][1];
    dp[u][0] = x * (dp[v][0] + dp[v][1]);
    dp[u][1] = y * (dp[v][0] + dp[v][1]);
    dp[u][1] = dp[u][1] + y * dp[v][0];
    dp[u][0] = dp[u][0] + x * (dp[v][0] + dp[v][1]) + y * dp[v][1] +
               expe(1ll * y.cnt * dp[v][1].cnt);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
  printf("%d\n", 2 * (dp[1][0] + dp[1][1]).sum % mod);
}

### Prompt

Please formulate 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 int mod = 998244353;
struct expe {
  int sum, cnt;
  expe(long long s = 0, long long c = 0) { sum = s % mod, cnt = c % mod; }
  expe operator+(expe x) { return expe(sum + x.sum, cnt + x.cnt); }
  expe operator*(expe x) {
    return expe(1ll * sum * x.cnt + 1ll * x.sum * cnt, 1ll * cnt * x.cnt);
  }
} dp[500050][2];
int n;
vector<int> adj[500050];
void dfs(int u, int pa = 0) {
  dp[u][1] = expe(0, 1);
  for (auto v : adj[u]) {
    if (v == pa) continue;
    dfs(v, u);
    expe x = dp[u][0], y = dp[u][1];
    dp[u][0] = x * (dp[v][0] + dp[v][1]);
    dp[u][1] = y * (dp[v][0] + dp[v][1]);
    dp[u][1] = dp[u][1] + y * dp[v][0];
    dp[u][0] = dp[u][0] + x * (dp[v][0] + dp[v][1]) + y * dp[v][1] +
               expe(1ll * y.cnt * dp[v][1].cnt);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
  printf("%d\n", 2 * (dp[1][0] + dp[1][1]).sum % mod);
}

```