#include <bits/stdc++.h>
using namespace std;
const int N = 500000;
const int mo = 998244353;
int n;
int dp[N + 5][2];
vector<int> G[N + 5];
int fpm(int x, int y) {
  int res = 1;
  for (; y > 0; y >>= 1) {
    if (y & 1) res = (long long)res * x % mo;
    x = (long long)x * x % mo;
  }
  return res;
}
int ans;
int dfs(int u, int f = 0) {
  int sz = 1;
  dp[u][0] = 1;
  for (auto v : G[u]) {
    if (v == f) continue;
    sz += dfs(v, u);
    dp[u][1] = ((long long)dp[u][1] * (dp[v][0] + dp[v][1]) % mo +
                (long long)dp[u][0] * dp[v][0]) %
               mo;
    dp[u][0] = (long long)dp[u][0] * dp[v][1] % mo;
  }
  ans = (ans + (long long)fpm(2, n - sz + 1) * dp[u][1]) % mo;
  dp[u][1] = (dp[u][1] * 2 % mo + dp[u][0]) % mo;
  return sz;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    static int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v), G[v].push_back(u);
  }
  dfs(1);
  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 N = 500000;
const int mo = 998244353;
int n;
int dp[N + 5][2];
vector<int> G[N + 5];
int fpm(int x, int y) {
  int res = 1;
  for (; y > 0; y >>= 1) {
    if (y & 1) res = (long long)res * x % mo;
    x = (long long)x * x % mo;
  }
  return res;
}
int ans;
int dfs(int u, int f = 0) {
  int sz = 1;
  dp[u][0] = 1;
  for (auto v : G[u]) {
    if (v == f) continue;
    sz += dfs(v, u);
    dp[u][1] = ((long long)dp[u][1] * (dp[v][0] + dp[v][1]) % mo +
                (long long)dp[u][0] * dp[v][0]) %
               mo;
    dp[u][0] = (long long)dp[u][0] * dp[v][1] % mo;
  }
  ans = (ans + (long long)fpm(2, n - sz + 1) * dp[u][1]) % mo;
  dp[u][1] = (dp[u][1] * 2 % mo + dp[u][0]) % mo;
  return sz;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    static int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v), G[v].push_back(u);
  }
  dfs(1);
  printf("%d\n", ans);
  return 0;
}

```