#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10, mod = 998244353, inv2 = (mod + 1) / 2;
int n;
vector<int> g[N];
long long ans, anpr = 1, f[N];
void dfs(int u, int fa) {
  long long &e = f[u] = 1;
  for (int v : g[u]) {
    if (v != fa) {
      dfs(v, u);
      e = e * (1 - (1 - f[v]) * inv2 % mod) % mod;
    }
  }
  e = 1 - e;
  ans = (ans + 2 * e) % mod;
}
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; ++i) {
    anpr = anpr * 2 % mod;
    scanf("%d%d", &u, &v);
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  dfs(1, 0);
  printf("%lld\n", (ans * anpr % mod + mod) % mod);
}

### Prompt

In cpp, your task is to solve 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;
const int N = 5e5 + 10, mod = 998244353, inv2 = (mod + 1) / 2;
int n;
vector<int> g[N];
long long ans, anpr = 1, f[N];
void dfs(int u, int fa) {
  long long &e = f[u] = 1;
  for (int v : g[u]) {
    if (v != fa) {
      dfs(v, u);
      e = e * (1 - (1 - f[v]) * inv2 % mod) % mod;
    }
  }
  e = 1 - e;
  ans = (ans + 2 * e) % mod;
}
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; ++i) {
    anpr = anpr * 2 % mod;
    scanf("%d%d", &u, &v);
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  dfs(1, 0);
  printf("%lld\n", (ans * anpr % mod + mod) % mod);
}

```