#include <bits/stdc++.h>
using namespace std;
const int BASE = 998244353;
int n;
vector<int> a[500500];
long long f[500500][2], ways[500500][2];
void dfs(int x, int par) {
  ways[x][0] = 1;
  for (int y : a[x])
    if (y != par) {
      dfs(y, x);
      long long oldF[2] = {f[x][0], f[x][1]},
                oldWays[2] = {ways[x][0], ways[x][1]};
      long long sumFY = (f[y][0] + f[y][1]) % BASE,
                sumWaysY = (ways[y][0] + ways[y][1]) % BASE;
      for (int i = 0; i < 2; i++) {
        f[x][i] = (oldWays[i] * sumFY + oldF[i] * sumWaysY) % BASE;
        ways[x][i] = oldWays[i] * sumWaysY % BASE;
      }
      f[x][1] = (f[x][1] + oldWays[0] * (f[y][0] + ways[y][0]) +
                 oldF[0] * ways[y][0]) %
                BASE;
      f[x][1] = (f[x][1] + oldWays[1] * sumFY + oldF[1] * sumWaysY) % BASE;
      ways[x][1] =
          (ways[x][1] + oldWays[0] * ways[y][0] + oldWays[1] * sumWaysY) % BASE;
      f[x][0] = (f[x][0] + oldWays[0] * f[y][1] + oldF[0] * ways[y][1]) % BASE;
      ways[x][0] = (ways[x][0] + oldWays[0] * ways[y][1]) % BASE;
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1, 0);
  cout << (f[1][0] + f[1][1]) * 2 % BASE << endl;
}

### Prompt

In cpp, your task is to solve 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 BASE = 998244353;
int n;
vector<int> a[500500];
long long f[500500][2], ways[500500][2];
void dfs(int x, int par) {
  ways[x][0] = 1;
  for (int y : a[x])
    if (y != par) {
      dfs(y, x);
      long long oldF[2] = {f[x][0], f[x][1]},
                oldWays[2] = {ways[x][0], ways[x][1]};
      long long sumFY = (f[y][0] + f[y][1]) % BASE,
                sumWaysY = (ways[y][0] + ways[y][1]) % BASE;
      for (int i = 0; i < 2; i++) {
        f[x][i] = (oldWays[i] * sumFY + oldF[i] * sumWaysY) % BASE;
        ways[x][i] = oldWays[i] * sumWaysY % BASE;
      }
      f[x][1] = (f[x][1] + oldWays[0] * (f[y][0] + ways[y][0]) +
                 oldF[0] * ways[y][0]) %
                BASE;
      f[x][1] = (f[x][1] + oldWays[1] * sumFY + oldF[1] * sumWaysY) % BASE;
      ways[x][1] =
          (ways[x][1] + oldWays[0] * ways[y][0] + oldWays[1] * sumWaysY) % BASE;
      f[x][0] = (f[x][0] + oldWays[0] * f[y][1] + oldF[0] * ways[y][1]) % BASE;
      ways[x][0] = (ways[x][0] + oldWays[0] * ways[y][1]) % BASE;
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1, 0);
  cout << (f[1][0] + f[1][1]) * 2 % BASE << endl;
}

```