#include <bits/stdc++.h>
using namespace std;
const int mo = 998244353;
int n, f[666666], res, sz[666666], mi[666666];
vector<int> e[666666];
void dfs(int u, int fa) {
  sz[u] = 1;
  f[u] = 1;
  for (int i = 0; i < e[u].size(); i++)
    if (e[u][i] != fa) {
      int v = e[u][i];
      dfs(v, u);
      sz[u] += sz[v];
      f[u] = 1LL * f[u] * (mi[sz[v]] - f[v] + mo) % mo;
    }
  res = (res + (1LL * mi[sz[u] - 1] - f[u]) * mi[n - sz[u]]) % mo;
}
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);
  }
  mi[0] = 1;
  for (int i = 1; i <= n; i++) mi[i] = mi[i - 1] * 2 % mo;
  dfs(1, 0);
  res = (res * 2 % mo + mo) % mo;
  cout << res;
  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 mo = 998244353;
int n, f[666666], res, sz[666666], mi[666666];
vector<int> e[666666];
void dfs(int u, int fa) {
  sz[u] = 1;
  f[u] = 1;
  for (int i = 0; i < e[u].size(); i++)
    if (e[u][i] != fa) {
      int v = e[u][i];
      dfs(v, u);
      sz[u] += sz[v];
      f[u] = 1LL * f[u] * (mi[sz[v]] - f[v] + mo) % mo;
    }
  res = (res + (1LL * mi[sz[u] - 1] - f[u]) * mi[n - sz[u]]) % mo;
}
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);
  }
  mi[0] = 1;
  for (int i = 1; i <= n; i++) mi[i] = mi[i - 1] * 2 % mo;
  dfs(1, 0);
  res = (res * 2 % mo + mo) % mo;
  cout << res;
  return 0;
}

```