#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
const int mod = 998244353;
int Head[N], vet[N << 1], Next[N << 1];
int siz[N], Pow[N], dp[N];
int n, u, v, edgenum, ans;
inline void addedge(int u, int v) {
  edgenum++;
  vet[edgenum] = v;
  Next[edgenum] = Head[u];
  Head[u] = edgenum;
}
void dfs(int u, int f) {
  int res = 1;
  siz[u] = 1;
  for (int e = Head[u]; e; e = Next[e]) {
    int v = vet[e];
    if (v == f) continue;
    dfs(v, u);
    siz[u] += siz[v];
    res = 1ll * res * (Pow[siz[v] - 1] + dp[v]) % mod;
  }
  dp[u] = (Pow[siz[u] - 1] - res + mod) % mod;
  ans = (ans + 1ll * Pow[n - siz[u]] * dp[u]) % mod;
}
int main() {
  scanf("%d", &n);
  Pow[0] = 1;
  for (int i = 1; i <= n; i++) Pow[i] = Pow[i - 1] * 2 % mod;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  dfs(1, 0);
  printf("%d\n", ans * 2 % mod);
  return 0;
}

### Prompt

Generate a CPP solution 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 = 5e5 + 5;
const int mod = 998244353;
int Head[N], vet[N << 1], Next[N << 1];
int siz[N], Pow[N], dp[N];
int n, u, v, edgenum, ans;
inline void addedge(int u, int v) {
  edgenum++;
  vet[edgenum] = v;
  Next[edgenum] = Head[u];
  Head[u] = edgenum;
}
void dfs(int u, int f) {
  int res = 1;
  siz[u] = 1;
  for (int e = Head[u]; e; e = Next[e]) {
    int v = vet[e];
    if (v == f) continue;
    dfs(v, u);
    siz[u] += siz[v];
    res = 1ll * res * (Pow[siz[v] - 1] + dp[v]) % mod;
  }
  dp[u] = (Pow[siz[u] - 1] - res + mod) % mod;
  ans = (ans + 1ll * Pow[n - siz[u]] * dp[u]) % mod;
}
int main() {
  scanf("%d", &n);
  Pow[0] = 1;
  for (int i = 1; i <= n; i++) Pow[i] = Pow[i - 1] * 2 % mod;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  dfs(1, 0);
  printf("%d\n", ans * 2 % mod);
  return 0;
}

```