#include <bits/stdc++.h>
using namespace std;
const int N = 501000, MOD = 998244353, inv = 499122177;
namespace {
int Add(int x, int y) { return (x += y) >= MOD ? x - MOD : x; }
int Dec(int x, int y) { return (x -= y) < 0 ? x + MOD : x; }
int Mul(int x, int y) { return 1LL * x * y % MOD; }
}  // namespace
int n, f[N], Ans;
vector<int> g[N];
void Dfs(int x, int fa) {
  f[x] = 1;
  for (int v : g[x]) {
    if (v == fa) continue;
    Dfs(v, x);
    f[x] = Mul(f[x], Dec(1, Mul(inv, Dec(1, f[v]))));
  }
  f[x] = Dec(1, f[x]);
  Ans = Add(Ans, f[x]);
}
int main(void) {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  Dfs(1, 0);
  for (int i = 1; i <= n; i++) Ans = Mul(Ans, 2);
  cout << Ans << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
const int N = 501000, MOD = 998244353, inv = 499122177;
namespace {
int Add(int x, int y) { return (x += y) >= MOD ? x - MOD : x; }
int Dec(int x, int y) { return (x -= y) < 0 ? x + MOD : x; }
int Mul(int x, int y) { return 1LL * x * y % MOD; }
}  // namespace
int n, f[N], Ans;
vector<int> g[N];
void Dfs(int x, int fa) {
  f[x] = 1;
  for (int v : g[x]) {
    if (v == fa) continue;
    Dfs(v, x);
    f[x] = Mul(f[x], Dec(1, Mul(inv, Dec(1, f[v]))));
  }
  f[x] = Dec(1, f[x]);
  Ans = Add(Ans, f[x]);
}
int main(void) {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  Dfs(1, 0);
  for (int i = 1; i <= n; i++) Ans = Mul(Ans, 2);
  cout << Ans << endl;
  return 0;
}

```