#include <bits/stdc++.h>
const int N = 510000, M = N * 2;
const long long mod = 998244353, inv2 = (mod + 1) / 2;
int n, E = 0;
int to[M], first[N], next[M];
int cnt = 0, id[N], f[N];
long long ans = 0;
long long PowerMod(long long a, int n, long long c = 1) {
  for (; n; n >>= 1, a = a * a % mod)
    if (n & 1) c = c * a % mod;
  return c;
}
inline void addedge(int u, int v) {
  to[++E] = v;
  next[E] = first[u];
  first[u] = E;
  to[++E] = u;
  next[E] = first[v];
  first[v] = E;
}
void dfs(int x, int px = 0) {
  int i, y;
  long long cur = 1;
  for (i = first[x]; i; i = next[i])
    if ((y = to[i]) != px) dfs(y, x), cur = cur * (1 + f[y]) % mod * inv2 % mod;
  ans += f[x] = mod + 1 - cur;
}
int main() {
  int i, u, v;
  scanf("%d", &n);
  for (i = 1; i < n; ++i) scanf("%d%d", &u, &v), addedge(u, v);
  dfs(1);
  ans %= mod;
  printf("%lld\n", PowerMod(2, n, ans + (ans >> 63 & 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>
const int N = 510000, M = N * 2;
const long long mod = 998244353, inv2 = (mod + 1) / 2;
int n, E = 0;
int to[M], first[N], next[M];
int cnt = 0, id[N], f[N];
long long ans = 0;
long long PowerMod(long long a, int n, long long c = 1) {
  for (; n; n >>= 1, a = a * a % mod)
    if (n & 1) c = c * a % mod;
  return c;
}
inline void addedge(int u, int v) {
  to[++E] = v;
  next[E] = first[u];
  first[u] = E;
  to[++E] = u;
  next[E] = first[v];
  first[v] = E;
}
void dfs(int x, int px = 0) {
  int i, y;
  long long cur = 1;
  for (i = first[x]; i; i = next[i])
    if ((y = to[i]) != px) dfs(y, x), cur = cur * (1 + f[y]) % mod * inv2 % mod;
  ans += f[x] = mod + 1 - cur;
}
int main() {
  int i, u, v;
  scanf("%d", &n);
  for (i = 1; i < n; ++i) scanf("%d%d", &u, &v), addedge(u, v);
  dfs(1);
  ans %= mod;
  printf("%lld\n", PowerMod(2, n, ans + (ans >> 63 & mod)));
  return 0;
}

```