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

### Prompt

Develop a solution in CPP to the problem described below:

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;
const int mod = 998244353;
vector<int> g[maxn];
int n;
int dp[2][maxn], sz[maxn];
int pw[maxn];
int ans;
void dfs(int u, int fa) {
  dp[0][u] = 1;
  sz[u] = 1;
  for (int v : g[u])
    if (v != fa) {
      dfs(v, u);
      sz[u] += sz[v];
      dp[1][u] = (2LL * dp[1][u] * (dp[0][v] + dp[1][v]) % mod +
                  1LL * dp[0][u] * dp[0][v] % mod) %
                 mod;
      dp[0][u] = 1LL * dp[0][u] * (2LL * dp[1][v] + dp[0][v]) % mod;
    }
  ans = (ans + 2LL * dp[1][u] * pw[n - sz[u]]) % mod;
}
int main() {
  pw[0] = 1;
  for (int i = 1; i < maxn; i++) pw[i] = pw[i - 1] * 2 % mod;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y), g[y].push_back(x);
  }
  dfs(1, -1);
  printf("%d\n", ans);
  return 0;
}

```