#include <bits/stdc++.h>
using namespace std;
const int Mod = 998244353, inv2 = (Mod + 1) >> 1;
const int MAX_N = 5e5 + 5;
vector<int> E[MAX_N];
int N, f[MAX_N];
void dfs(int x, int fa) {
  f[x] = 1;
  for (int v : E[x]) {
    if (v == fa) continue;
    dfs(v, x);
    f[x] = f[x] * (1 - 1ll * (1 - f[v] + Mod) * inv2 % Mod + Mod) % Mod;
  }
  f[x] = (Mod + 1 - f[x]) % Mod;
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i < N; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    E[u].push_back(v), E[v].push_back(u);
  }
  dfs(1, 0);
  int ans = 0;
  for (int i = 1; i <= N; i++) ans = (ans + f[i]) % Mod;
  for (int i = 1; i <= N; i++) ans = (ans + ans) % Mod;
  printf("%d\n", ans);
  return 0;
}

### 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 Mod = 998244353, inv2 = (Mod + 1) >> 1;
const int MAX_N = 5e5 + 5;
vector<int> E[MAX_N];
int N, f[MAX_N];
void dfs(int x, int fa) {
  f[x] = 1;
  for (int v : E[x]) {
    if (v == fa) continue;
    dfs(v, x);
    f[x] = f[x] * (1 - 1ll * (1 - f[v] + Mod) * inv2 % Mod + Mod) % Mod;
  }
  f[x] = (Mod + 1 - f[x]) % Mod;
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i < N; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    E[u].push_back(v), E[v].push_back(u);
  }
  dfs(1, 0);
  int ans = 0;
  for (int i = 1; i <= N; i++) ans = (ans + f[i]) % Mod;
  for (int i = 1; i <= N; i++) ans = (ans + ans) % Mod;
  printf("%d\n", ans);
  return 0;
}

```