#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)5e5 + 1, mod = 998244353;
int n, f[maxn][2], sz[maxn], pw2[maxn];
vector<int> e[maxn];
inline void mod_inc(int &x, int y) { (x += y) >= mod && (x -= mod); }
void dfs(int u, int fa) {
  ++sz[u];
  f[u][0] = 1;
  f[u][1] = 0;
  for (int v : e[u]) {
    if (v == fa) continue;
    dfs(v, u);
    sz[u] += sz[v];
    int sum = f[v][0];
    mod_inc(sum, f[v][1]);
    int nxt[2] = {};
    nxt[0] = (long long)f[u][0] * (sum + f[v][1]) % mod;
    nxt[1] =
        ((long long)f[u][0] * f[v][0] + (long long)f[u][1] * (sum + sum)) % mod;
    f[u][0] = nxt[0];
    f[u][1] = nxt[1];
  }
}
int main() {
  pw2[0] = 1;
  for (int i = 1; i < maxn; ++i) {
    pw2[i] = pw2[i - 1];
    mod_inc(pw2[i], pw2[i - 1]);
  }
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, -1);
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    ans = (ans + (long long)pw2[n - sz[i]] * f[i][1]) % mod;
  mod_inc(ans, ans);
  printf("%d\n", ans);
  return 0;
}

### 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;
const int maxn = (int)5e5 + 1, mod = 998244353;
int n, f[maxn][2], sz[maxn], pw2[maxn];
vector<int> e[maxn];
inline void mod_inc(int &x, int y) { (x += y) >= mod && (x -= mod); }
void dfs(int u, int fa) {
  ++sz[u];
  f[u][0] = 1;
  f[u][1] = 0;
  for (int v : e[u]) {
    if (v == fa) continue;
    dfs(v, u);
    sz[u] += sz[v];
    int sum = f[v][0];
    mod_inc(sum, f[v][1]);
    int nxt[2] = {};
    nxt[0] = (long long)f[u][0] * (sum + f[v][1]) % mod;
    nxt[1] =
        ((long long)f[u][0] * f[v][0] + (long long)f[u][1] * (sum + sum)) % mod;
    f[u][0] = nxt[0];
    f[u][1] = nxt[1];
  }
}
int main() {
  pw2[0] = 1;
  for (int i = 1; i < maxn; ++i) {
    pw2[i] = pw2[i - 1];
    mod_inc(pw2[i], pw2[i - 1]);
  }
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, -1);
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    ans = (ans + (long long)pw2[n - sz[i]] * f[i][1]) % mod;
  mod_inc(ans, ans);
  printf("%d\n", ans);
  return 0;
}

```