#include <bits/stdc++.h>
using namespace std;
const int N = 500005, mod = 998244353, inv2 = (mod + 1) / 2;
int n, h[N], cnt;
long long f[N], ans;
struct edge {
  int v, n;
} e[N << 1];
int getint() {
  char ch;
  while (!isdigit(ch = getchar()))
    ;
  int x = ch - 48;
  while (isdigit(ch = getchar())) x = x * 10 + ch - 48;
  return x;
}
void addedge(int u, int v) {
  e[cnt] = (edge){v, h[u]}, h[u] = cnt++;
  e[cnt] = (edge){u, h[v]}, h[v] = cnt++;
}
void dfs(int x, int fa) {
  f[x] = 1;
  for (int i = h[x]; i != -1; i = e[i].n)
    if (e[i].v != fa) {
      dfs(e[i].v, x);
      f[x] = (1 - inv2 * (1 - f[e[i].v])) % mod * f[x] % mod;
    }
  f[x] = (1 - f[x] + mod) % mod;
}
int main() {
  n = getint();
  memset(h, -1, sizeof(h));
  for (int i = 1; i <= n - 1; ++i) addedge(getint(), getint());
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) ans = (ans + f[i]) % mod;
  long long pw = 1;
  for (int i = 1; i <= n; ++i) pw = pw * 2 % mod;
  printf("%lld\n", ans * pw % 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 N = 500005, mod = 998244353, inv2 = (mod + 1) / 2;
int n, h[N], cnt;
long long f[N], ans;
struct edge {
  int v, n;
} e[N << 1];
int getint() {
  char ch;
  while (!isdigit(ch = getchar()))
    ;
  int x = ch - 48;
  while (isdigit(ch = getchar())) x = x * 10 + ch - 48;
  return x;
}
void addedge(int u, int v) {
  e[cnt] = (edge){v, h[u]}, h[u] = cnt++;
  e[cnt] = (edge){u, h[v]}, h[v] = cnt++;
}
void dfs(int x, int fa) {
  f[x] = 1;
  for (int i = h[x]; i != -1; i = e[i].n)
    if (e[i].v != fa) {
      dfs(e[i].v, x);
      f[x] = (1 - inv2 * (1 - f[e[i].v])) % mod * f[x] % mod;
    }
  f[x] = (1 - f[x] + mod) % mod;
}
int main() {
  n = getint();
  memset(h, -1, sizeof(h));
  for (int i = 1; i <= n - 1; ++i) addedge(getint(), getint());
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) ans = (ans + f[i]) % mod;
  long long pw = 1;
  for (int i = 1; i <= n; ++i) pw = pw * 2 % mod;
  printf("%lld\n", ans * pw % mod);
  return 0;
}

```