#include <bits/stdc++.h>
const int maxn = 500010, mod = 998244353;
int n;
struct edge {
  int to;
  edge* next;
} E[maxn * 2], *ne = E, *fir[maxn];
void link(int u, int v) {
  *ne = (edge){v, fir[u]};
  fir[u] = ne++;
}
int pw[maxn], f[maxn], sz[maxn];
int dfs(int i, int fa = 0) {
  int sum = 0, w = sz[i] = 1;
  for (edge* e = fir[i]; e; e = e->next)
    if (e->to != fa) {
      (sum += dfs(e->to, i)) %= mod;
      sz[i] += sz[e->to];
      w = 1ll * w * (f[e->to] + pw[sz[e->to] - 1]) % mod;
    }
  f[i] = (pw[sz[i] - 1] - w + mod) % mod;
  return (sum + 1ll * f[i] * pw[n - sz[i]]) % mod;
}
int main() {
  scanf("%d", &n);
  for (int i = *pw = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    link(u, v);
    link(v, u);
    pw[i] = pw[i - 1] * 2 % mod;
  }
  printf("%d\n", dfs(1) * 2 % mod);
}

### 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>
const int maxn = 500010, mod = 998244353;
int n;
struct edge {
  int to;
  edge* next;
} E[maxn * 2], *ne = E, *fir[maxn];
void link(int u, int v) {
  *ne = (edge){v, fir[u]};
  fir[u] = ne++;
}
int pw[maxn], f[maxn], sz[maxn];
int dfs(int i, int fa = 0) {
  int sum = 0, w = sz[i] = 1;
  for (edge* e = fir[i]; e; e = e->next)
    if (e->to != fa) {
      (sum += dfs(e->to, i)) %= mod;
      sz[i] += sz[e->to];
      w = 1ll * w * (f[e->to] + pw[sz[e->to] - 1]) % mod;
    }
  f[i] = (pw[sz[i] - 1] - w + mod) % mod;
  return (sum + 1ll * f[i] * pw[n - sz[i]]) % mod;
}
int main() {
  scanf("%d", &n);
  for (int i = *pw = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    link(u, v);
    link(v, u);
    pw[i] = pw[i - 1] * 2 % mod;
  }
  printf("%d\n", dfs(1) * 2 % mod);
}

```