#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int n, ans, pw[3000000], x, y, tot, head[3000000], sz[3000000], f[3000000],
    g[3000000];
struct edge {
  int v, nxt;
} e[3000000];
void add(int x, int y) {
  e[++tot].v = y;
  e[tot].nxt = head[x];
  head[x] = tot;
}
void dfs(int u, int fa) {
  f[u] = 1;
  sz[u] = 1;
  for (int i = head[u], v; i; i = e[i].nxt)
    if ((v = e[i].v) != fa) {
      dfs(v, u);
      sz[u] += sz[v];
      f[u] = f[u] * ((long long)g[v] + pw[sz[v] - 1]) % mod;
    }
  g[u] = (pw[sz[u] - 1] - f[u] + mod) % mod;
  ans = (ans + (long long)g[u] * pw[n - sz[u]]) % mod;
}
int main() {
  scanf("%d", &n);
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = pw[i - 1] * 2 % mod;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs(1, 0);
  printf("%d\n", ans * 2 % mod);
  return 0;
}

### Prompt

Please create a solution in Cpp 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 mod = 998244353;
int n, ans, pw[3000000], x, y, tot, head[3000000], sz[3000000], f[3000000],
    g[3000000];
struct edge {
  int v, nxt;
} e[3000000];
void add(int x, int y) {
  e[++tot].v = y;
  e[tot].nxt = head[x];
  head[x] = tot;
}
void dfs(int u, int fa) {
  f[u] = 1;
  sz[u] = 1;
  for (int i = head[u], v; i; i = e[i].nxt)
    if ((v = e[i].v) != fa) {
      dfs(v, u);
      sz[u] += sz[v];
      f[u] = f[u] * ((long long)g[v] + pw[sz[v] - 1]) % mod;
    }
  g[u] = (pw[sz[u] - 1] - f[u] + mod) % mod;
  ans = (ans + (long long)g[u] * pw[n - sz[u]]) % mod;
}
int main() {
  scanf("%d", &n);
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = pw[i - 1] * 2 % mod;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs(1, 0);
  printf("%d\n", ans * 2 % mod);
  return 0;
}

```