#include <bits/stdc++.h>
using namespace std;
int n, x, y, fa[1000000];
long long F[1000000], f[1000000], G[1000000], g[1000000];
vector<int> p[1000000];
void dp(int x) {
  F[x] = 0;
  f[x] = 1;
  for (auto i : p[x])
    if (i != fa[x]) {
      fa[i] = x;
      dp(i);
      G[x] = (G[x] * (f[i] + g[i]) * 2 + F[x] * f[i] +
              F[i] * (g[x] * 2 + f[x]) + G[i] * g[x] * 2 + f[x] * f[i]) %
             998244353;
      g[x] = (g[x] * (f[i] + g[i]) * 2 + f[x] * f[i]) % 998244353;
      F[x] = (F[x] * (f[i] + g[i] * 2) + f[x] * (F[i] + G[i] * 2)) % 998244353;
      f[x] = f[x] * (f[i] + g[i] * 2) % 998244353;
    }
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n - 1); i++) {
    scanf("%d%d", &x, &y);
    p[x].push_back(y);
    p[y].push_back(x);
  }
  dp(1);
  printf("%lld\n", (F[1] + G[1]) * 2 % 998244353);
}

### Prompt

Please provide a cpp coded solution 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;
int n, x, y, fa[1000000];
long long F[1000000], f[1000000], G[1000000], g[1000000];
vector<int> p[1000000];
void dp(int x) {
  F[x] = 0;
  f[x] = 1;
  for (auto i : p[x])
    if (i != fa[x]) {
      fa[i] = x;
      dp(i);
      G[x] = (G[x] * (f[i] + g[i]) * 2 + F[x] * f[i] +
              F[i] * (g[x] * 2 + f[x]) + G[i] * g[x] * 2 + f[x] * f[i]) %
             998244353;
      g[x] = (g[x] * (f[i] + g[i]) * 2 + f[x] * f[i]) % 998244353;
      F[x] = (F[x] * (f[i] + g[i] * 2) + f[x] * (F[i] + G[i] * 2)) % 998244353;
      f[x] = f[x] * (f[i] + g[i] * 2) % 998244353;
    }
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n - 1); i++) {
    scanf("%d%d", &x, &y);
    p[x].push_back(y);
    p[y].push_back(x);
  }
  dp(1);
  printf("%lld\n", (F[1] + G[1]) * 2 % 998244353);
}

```