#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return {i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const int N = 5e5 + 7;
int n;
vector<int> G[N];
int par[N];
int num[N];
int ile[N];
int pot[N];
int roz[N];
int dp[N];
const int mod = 998244353;
void dfs0(int v, int p) {
  par[v] = p;
  int nn = 0;
  roz[v] = 1;
  for (int u : G[v]) {
    if (u == p) continue;
    num[u] = ++nn;
    dfs0(u, v);
    roz[v] += roz[u];
  }
  ile[v] = G[v].size();
  if (p) ile[v]--;
}
void dfs1(int v, int p) {
  for (int u : G[v]) {
    if (u == p) continue;
    dfs1(u, v);
  }
  long long wynik = 1;
  for (int u : G[v]) {
    if (u == p) continue;
    int pom = (pot[roz[u] - 1] + dp[u]) % mod;
    wynik = (wynik * pom) % mod;
  }
  wynik = mod - wynik;
  wynik += pot[roz[v] - 1];
  dp[v] = wynik % mod;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs0(1, 0);
  pot[0] = 1;
  for (int i = 1; i <= n; ++i) pot[i] = (2 * pot[i - 1]) % mod;
  dfs1(1, 0);
  long long wynik = 0;
  for (int i = 1; i <= n; ++i) {
    long long pom = dp[i];
    pom *= pot[n - roz[i]];
    wynik += pom % mod;
  }
  wynik *= 2;
  wynik %= mod;
  printf("%lld\n", wynik);
  return 0;
}

### 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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return {i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const int N = 5e5 + 7;
int n;
vector<int> G[N];
int par[N];
int num[N];
int ile[N];
int pot[N];
int roz[N];
int dp[N];
const int mod = 998244353;
void dfs0(int v, int p) {
  par[v] = p;
  int nn = 0;
  roz[v] = 1;
  for (int u : G[v]) {
    if (u == p) continue;
    num[u] = ++nn;
    dfs0(u, v);
    roz[v] += roz[u];
  }
  ile[v] = G[v].size();
  if (p) ile[v]--;
}
void dfs1(int v, int p) {
  for (int u : G[v]) {
    if (u == p) continue;
    dfs1(u, v);
  }
  long long wynik = 1;
  for (int u : G[v]) {
    if (u == p) continue;
    int pom = (pot[roz[u] - 1] + dp[u]) % mod;
    wynik = (wynik * pom) % mod;
  }
  wynik = mod - wynik;
  wynik += pot[roz[v] - 1];
  dp[v] = wynik % mod;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs0(1, 0);
  pot[0] = 1;
  for (int i = 1; i <= n; ++i) pot[i] = (2 * pot[i - 1]) % mod;
  dfs1(1, 0);
  long long wynik = 0;
  for (int i = 1; i <= n; ++i) {
    long long pom = dp[i];
    pom *= pot[n - roz[i]];
    wynik += pom % mod;
  }
  wynik *= 2;
  wynik %= mod;
  printf("%lld\n", wynik);
  return 0;
}

```