#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10;
int add(int x, int y) {
  x += y;
  if (x >= 998244353) x -= 998244353;
  return x;
}
int sub(int x, int y) {
  if (x < y) x += 998244353;
  x -= y;
  return x;
}
int mul(int x, int y) { return (long long)x * y % 998244353; }
int qpow(int x, int y, int p = 998244353) {
  int res = 1;
  while (y) {
    if (y & 1) res = mul(res, x);
    x = mul(x, x);
    y >>= 1;
  }
  return res;
}
vector<int> G[N];
int f[N], ans;
void dfs(int x, int fa) {
  f[x] = 1;
  for (int j = 0; j < G[x].size(); j++)
    if (G[x][j] != fa) {
      int v = G[x][j];
      dfs(v, x);
      f[x] = mul(f[x], add(1 + 998244353 - f[v], mul(f[v], 499122177)));
    }
  ans = add(ans, 998244353 + 1 - f[x]);
}
int n;
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  ans = mul(ans, qpow(2, n));
  cout << 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 N = 5e5 + 10;
int add(int x, int y) {
  x += y;
  if (x >= 998244353) x -= 998244353;
  return x;
}
int sub(int x, int y) {
  if (x < y) x += 998244353;
  x -= y;
  return x;
}
int mul(int x, int y) { return (long long)x * y % 998244353; }
int qpow(int x, int y, int p = 998244353) {
  int res = 1;
  while (y) {
    if (y & 1) res = mul(res, x);
    x = mul(x, x);
    y >>= 1;
  }
  return res;
}
vector<int> G[N];
int f[N], ans;
void dfs(int x, int fa) {
  f[x] = 1;
  for (int j = 0; j < G[x].size(); j++)
    if (G[x][j] != fa) {
      int v = G[x][j];
      dfs(v, x);
      f[x] = mul(f[x], add(1 + 998244353 - f[v], mul(f[v], 499122177)));
    }
  ans = add(ans, 998244353 + 1 - f[x]);
}
int n;
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  ans = mul(ans, qpow(2, n));
  cout << ans;
  return 0;
}

```