#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10, P = 998244353;
int n, u, v;
vector<int> G[N];
inline unsigned norm(unsigned x) { return x >= P ? x - P : x; }
struct Z {
  unsigned x;
  Z(unsigned _x = 0) : x(_x) {}
} inv2(P + 1 >> 1), dp[N], ans;
inline Z operator+(const Z a, const Z b) { return norm(a.x + b.x); }
inline Z operator-(const Z a, const Z b) { return norm(a.x + P - b.x); }
inline Z operator*(const Z a, const Z b) {
  return static_cast<unsigned long long>(a.x) * b.x % P;
}
inline Z operator-(const Z a) { return a.x ? P - a.x : 0; }
inline Z operator+=(Z &a, const Z b) { return a = a + b; }
inline Z operator-=(Z &a, const Z b) { return a = a - b; }
inline Z operator*=(Z &a, const Z b) { return a = a * b; }
Z ksm(Z a, int b) {
  Z ans(1);
  for (; b; b >>= 1, a *= a)
    if (b & 1) ans *= a;
  return ans;
}
void dfs(int u, int p) {
  dp[u] = 1;
  for (int &v : G[u])
    if (v ^ p) dfs(v, u), dp[u] *= inv2 * (dp[v] + 1);
  dp[u] = 1 - dp[u];
  ans += dp[u];
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n - 1); ++i)
    scanf("%d%d", &u, &v), G[u].push_back(v), G[v].push_back(u);
  dfs(1, 0);
  printf("%u\n", (ans * ksm(2, n)).x);
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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>
using namespace std;
const int N = 5e5 + 10, P = 998244353;
int n, u, v;
vector<int> G[N];
inline unsigned norm(unsigned x) { return x >= P ? x - P : x; }
struct Z {
  unsigned x;
  Z(unsigned _x = 0) : x(_x) {}
} inv2(P + 1 >> 1), dp[N], ans;
inline Z operator+(const Z a, const Z b) { return norm(a.x + b.x); }
inline Z operator-(const Z a, const Z b) { return norm(a.x + P - b.x); }
inline Z operator*(const Z a, const Z b) {
  return static_cast<unsigned long long>(a.x) * b.x % P;
}
inline Z operator-(const Z a) { return a.x ? P - a.x : 0; }
inline Z operator+=(Z &a, const Z b) { return a = a + b; }
inline Z operator-=(Z &a, const Z b) { return a = a - b; }
inline Z operator*=(Z &a, const Z b) { return a = a * b; }
Z ksm(Z a, int b) {
  Z ans(1);
  for (; b; b >>= 1, a *= a)
    if (b & 1) ans *= a;
  return ans;
}
void dfs(int u, int p) {
  dp[u] = 1;
  for (int &v : G[u])
    if (v ^ p) dfs(v, u), dp[u] *= inv2 * (dp[v] + 1);
  dp[u] = 1 - dp[u];
  ans += dp[u];
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n - 1); ++i)
    scanf("%d%d", &u, &v), G[u].push_back(v), G[v].push_back(u);
  dfs(1, 0);
  printf("%u\n", (ans * ksm(2, n)).x);
  return 0;
}

```