#include <bits/stdc++.h>
using namespace std;
const int maxn = 500005, mod = 998244353, inv2 = (mod + 1) / 2;
int n, pw, ans;
int f[maxn];
vector<int> v[maxn];
void dfs(int x, int last) {
  f[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    int y = v[x][i];
    if (y == last) continue;
    dfs(y, x);
    f[x] = 1ll * f[x] * (inv2 + 1ll * (1 - f[y] + mod) * inv2 % mod) % mod;
  }
  ans = (ans + (1 - f[x] + mod) % mod) % mod;
}
int main() {
  scanf("%d", &n), pw = 1;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y), v[y].push_back(x);
    pw = (pw + pw) % mod;
  }
  dfs(1, 0);
  printf("%d\n", 2ll * ans * pw % mod);
  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 maxn = 500005, mod = 998244353, inv2 = (mod + 1) / 2;
int n, pw, ans;
int f[maxn];
vector<int> v[maxn];
void dfs(int x, int last) {
  f[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    int y = v[x][i];
    if (y == last) continue;
    dfs(y, x);
    f[x] = 1ll * f[x] * (inv2 + 1ll * (1 - f[y] + mod) * inv2 % mod) % mod;
  }
  ans = (ans + (1 - f[x] + mod) % mod) % mod;
}
int main() {
  scanf("%d", &n), pw = 1;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y), v[y].push_back(x);
    pw = (pw + pw) % mod;
  }
  dfs(1, 0);
  printf("%d\n", 2ll * ans * pw % mod);
  return 0;
}

```