#include <bits/stdc++.h>
using namespace std;
long long gi() {
  char cc = getchar();
  long long cn = 0, flus = 1;
  while (cc < '0' || cc > '9') {
    if (cc == '-') flus = -flus;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') cn = cn * 10 + cc - '0', cc = getchar();
  return cn * flus;
}
const long long N = 5e5 + 5;
const long long P = 998244353;
long long n, Ans, sz[N], fac[N], f[N][2];
vector<long long> G[N];
void add(long long x, long long y) { G[x].push_back(y), G[y].push_back(x); }
void dfs(long long x, long long fa) {
  sz[x] = 1, f[x][0] = 1;
  for (long long v : G[x]) {
    if (v == fa) continue;
    dfs(v, x), sz[x] += sz[v];
    f[x][0] = f[x][0] * (f[v][0] + f[v][1] + f[v][1]) % P;
  }
  f[x][1] = (fac[sz[x] - 1] - f[x][0] + P) % P;
  Ans = (Ans + f[x][1] * fac[n - sz[x]]) % P;
}
signed main() {
  n = gi();
  long long x, y;
  for (register long long i = (2); i <= (n); ++i) x = gi(), y = gi(), add(x, y);
  fac[0] = 1;
  for (register long long i = (1); i <= (n); ++i) fac[i] = fac[i - 1] * 2 % P;
  dfs(1, 1);
  cout << Ans * 2 % P << endl;
  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;
long long gi() {
  char cc = getchar();
  long long cn = 0, flus = 1;
  while (cc < '0' || cc > '9') {
    if (cc == '-') flus = -flus;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') cn = cn * 10 + cc - '0', cc = getchar();
  return cn * flus;
}
const long long N = 5e5 + 5;
const long long P = 998244353;
long long n, Ans, sz[N], fac[N], f[N][2];
vector<long long> G[N];
void add(long long x, long long y) { G[x].push_back(y), G[y].push_back(x); }
void dfs(long long x, long long fa) {
  sz[x] = 1, f[x][0] = 1;
  for (long long v : G[x]) {
    if (v == fa) continue;
    dfs(v, x), sz[x] += sz[v];
    f[x][0] = f[x][0] * (f[v][0] + f[v][1] + f[v][1]) % P;
  }
  f[x][1] = (fac[sz[x] - 1] - f[x][0] + P) % P;
  Ans = (Ans + f[x][1] * fac[n - sz[x]]) % P;
}
signed main() {
  n = gi();
  long long x, y;
  for (register long long i = (2); i <= (n); ++i) x = gi(), y = gi(), add(x, y);
  fac[0] = 1;
  for (register long long i = (1); i <= (n); ++i) fac[i] = fac[i - 1] * 2 % P;
  dfs(1, 1);
  cout << Ans * 2 % P << endl;
  return 0;
}

```