#include <bits/stdc++.h>
const int N = 500005;
const int mod = 998244353;
const int i2 = mod + 1 >> 1;
using LL = long long;
int x, y, head[N], tot, n;
struct edge {
  int to, nxt;
} e[N << 1];
void addedge(int x, int y) {
  e[++tot] = (edge){y, head[x]};
  head[x] = tot;
  e[++tot] = (edge){x, head[y]};
  head[y] = tot;
}
int pow(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = static_cast<LL>(x) * x % mod)
    if (y & 1) ans = static_cast<LL>(ans) * x % mod;
  return ans;
}
inline void reduce(int &x) { x += x >> 31 & mod; }
int ans = 0;
int dfs(int x, int f = 0) {
  int ans = 1;
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != f)
      ans = static_cast<LL>(ans) * (1 + dfs(e[i].to, x)) % mod * i2 % mod;
  reduce(ans = 1 - ans);
  reduce(::ans += ans - mod);
  return ans;
}
int main() {
  std::scanf("%d", &n);
  for (int i = 1; i < n; i++) std::scanf("%d%d", &x, &y), addedge(x, y);
  dfs(1);
  std::printf("%lld\n", static_cast<LL>(ans) * pow(2, n) % mod);
  return 0;
}

### Prompt

Create a solution in cpp for 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>
const int N = 500005;
const int mod = 998244353;
const int i2 = mod + 1 >> 1;
using LL = long long;
int x, y, head[N], tot, n;
struct edge {
  int to, nxt;
} e[N << 1];
void addedge(int x, int y) {
  e[++tot] = (edge){y, head[x]};
  head[x] = tot;
  e[++tot] = (edge){x, head[y]};
  head[y] = tot;
}
int pow(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = static_cast<LL>(x) * x % mod)
    if (y & 1) ans = static_cast<LL>(ans) * x % mod;
  return ans;
}
inline void reduce(int &x) { x += x >> 31 & mod; }
int ans = 0;
int dfs(int x, int f = 0) {
  int ans = 1;
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != f)
      ans = static_cast<LL>(ans) * (1 + dfs(e[i].to, x)) % mod * i2 % mod;
  reduce(ans = 1 - ans);
  reduce(::ans += ans - mod);
  return ans;
}
int main() {
  std::scanf("%d", &n);
  for (int i = 1; i < n; i++) std::scanf("%d%d", &x, &y), addedge(x, y);
  dfs(1);
  std::printf("%lld\n", static_cast<LL>(ans) * pow(2, n) % mod);
  return 0;
}

```