#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 10, mod = 998244353, INV = 499122177;
vector<int> v[maxn];
int dp[maxn];
int ans;
void dfs(int u, int par = -1) {
  int &p = dp[u] = 1;
  for (int y : v[u]) {
    if (y != par) {
      dfs(y, u);
      ans = (1ll * ans + 1ll * p * dp[y] % mod * INV) % mod;
      p = 1ll * p * (2 - dp[y]) % mod * INV % mod;
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1);
  for (int i = 0; i < n; i++) ans = 1ll * ans * 2 % mod;
  if (ans < 0) ans += mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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 = 5e5 + 10, mod = 998244353, INV = 499122177;
vector<int> v[maxn];
int dp[maxn];
int ans;
void dfs(int u, int par = -1) {
  int &p = dp[u] = 1;
  for (int y : v[u]) {
    if (y != par) {
      dfs(y, u);
      ans = (1ll * ans + 1ll * p * dp[y] % mod * INV) % mod;
      p = 1ll * p * (2 - dp[y]) % mod * INV % mod;
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1);
  for (int i = 0; i < n; i++) ans = 1ll * ans * 2 % mod;
  if (ans < 0) ans += mod;
  printf("%d\n", ans);
  return 0;
}

```