#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> g[500050];
int sz[500050];
void calc_sz(int u, int pa = 0) {
  sz[u] = 1;
  for (int v : g[u]) {
    if (v == pa) continue;
    calc_sz(v, u);
    sz[u] += sz[v];
  }
}
int p[500050];
const int mod = 998244353;
int s[500050], a[500050];
void dfs(int u, int pa = 0) {
  a[u] = 1;
  for (int v : g[u]) {
    if (v == pa) continue;
    dfs(v, u);
    s[u] = (s[u] + 1ll * s[v] * p[sz[u] - sz[v]]) % mod;
    a[u] = 1ll * a[u] * (a[v] + p[sz[v] - 1]) % mod;
  }
  a[u] = (p[sz[u] - 1] - a[u]) % mod;
  s[u] = (s[u] + a[u]) % mod;
}
int main() {
  cin >> n;
  p[0] = 1;
  for (int i = 1; i <= n; i++) p[i] = 2 * p[i - 1] % mod;
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  calc_sz(1);
  dfs(1);
  if (s[1] < 0) s[1] += mod;
  s[1] = 2 * s[1] % mod;
  cout << s[1];
}

### Prompt

Create a solution in Cpp for 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;
int n;
vector<int> g[500050];
int sz[500050];
void calc_sz(int u, int pa = 0) {
  sz[u] = 1;
  for (int v : g[u]) {
    if (v == pa) continue;
    calc_sz(v, u);
    sz[u] += sz[v];
  }
}
int p[500050];
const int mod = 998244353;
int s[500050], a[500050];
void dfs(int u, int pa = 0) {
  a[u] = 1;
  for (int v : g[u]) {
    if (v == pa) continue;
    dfs(v, u);
    s[u] = (s[u] + 1ll * s[v] * p[sz[u] - sz[v]]) % mod;
    a[u] = 1ll * a[u] * (a[v] + p[sz[v] - 1]) % mod;
  }
  a[u] = (p[sz[u] - 1] - a[u]) % mod;
  s[u] = (s[u] + a[u]) % mod;
}
int main() {
  cin >> n;
  p[0] = 1;
  for (int i = 1; i <= n; i++) p[i] = 2 * p[i - 1] % mod;
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  calc_sz(1);
  dfs(1);
  if (s[1] < 0) s[1] += mod;
  s[1] = 2 * s[1] % mod;
  cout << s[1];
}

```