#include <bits/stdc++.h>
const long long mod = 998244353;
const long long inv2 = (mod + 1) / 2;
const int maxn = 5E+5 + 5;
int n;
long long ans, dp[maxn];
std::vector<int> to[maxn];
inline void DFS(int u, int fa) {
  long long prod = 1;
  for (int v : to[u]) {
    if (v == fa) continue;
    DFS(v, u);
    prod = prod * (1 + dp[v]) % mod * inv2 % mod;
  }
  dp[u] = (1 - prod) % mod;
  (ans += dp[u]) %= mod;
}
int main() {
  scanf("%d", &n);
  long long pw = 2;
  for (int i = 1, u, v; i < n; ++i) {
    scanf("%d%d", &u, &v);
    to[u].push_back(v);
    to[v].push_back(u);
    pw = pw * 2 % mod;
  }
  DFS(1, 0);
  printf("%lld", (ans + mod) % mod * pw % mod);
}

### Prompt

Please create a solution in CPP to 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>
const long long mod = 998244353;
const long long inv2 = (mod + 1) / 2;
const int maxn = 5E+5 + 5;
int n;
long long ans, dp[maxn];
std::vector<int> to[maxn];
inline void DFS(int u, int fa) {
  long long prod = 1;
  for (int v : to[u]) {
    if (v == fa) continue;
    DFS(v, u);
    prod = prod * (1 + dp[v]) % mod * inv2 % mod;
  }
  dp[u] = (1 - prod) % mod;
  (ans += dp[u]) %= mod;
}
int main() {
  scanf("%d", &n);
  long long pw = 2;
  for (int i = 1, u, v; i < n; ++i) {
    scanf("%d%d", &u, &v);
    to[u].push_back(v);
    to[v].push_back(u);
    pw = pw * 2 % mod;
  }
  DFS(1, 0);
  printf("%lld", (ans + mod) % mod * pw % mod);
}

```