#include <bits/stdc++.h>
using namespace std;
int n, a[500005], siz[500005], f[500005];
vector<int> to[500005];
void dfs(int x, int p) {
  siz[x] = 1;
  long long tmp = 1;
  for (int y : to[x]) {
    if (y == p) continue;
    dfs(y, x);
    siz[x] += siz[y];
    tmp = tmp * (a[siz[y] - 1] + f[y]) % 998244353;
  }
  f[x] = (a[siz[x] - 1] - tmp + 998244353) % 998244353;
}
int main() {
  scanf("%d", &n);
  a[0] = 1;
  for (int i = 1; i < n; ++i) a[i] = a[i - 1] * 2LL % 998244353;
  for (int i = 1, x, y; i < n; ++i) {
    scanf("%d%d", &x, &y);
    to[x].push_back(y);
    to[y].push_back(x);
  }
  dfs(1, 0);
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    ans = (ans + 2LL * f[i] * a[n - siz[i]] % 998244353) % 998244353;
  printf("%d\n", ans);
}

### 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;
int n, a[500005], siz[500005], f[500005];
vector<int> to[500005];
void dfs(int x, int p) {
  siz[x] = 1;
  long long tmp = 1;
  for (int y : to[x]) {
    if (y == p) continue;
    dfs(y, x);
    siz[x] += siz[y];
    tmp = tmp * (a[siz[y] - 1] + f[y]) % 998244353;
  }
  f[x] = (a[siz[x] - 1] - tmp + 998244353) % 998244353;
}
int main() {
  scanf("%d", &n);
  a[0] = 1;
  for (int i = 1; i < n; ++i) a[i] = a[i - 1] * 2LL % 998244353;
  for (int i = 1, x, y; i < n; ++i) {
    scanf("%d%d", &x, &y);
    to[x].push_back(y);
    to[y].push_back(x);
  }
  dfs(1, 0);
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    ans = (ans + 2LL * f[i] * a[n - siz[i]] % 998244353) % 998244353;
  printf("%d\n", ans);
}

```