#include <bits/stdc++.h>
using namespace std;
int last[2000007], other[2000007], pre[2000007], cnt = 0;
int n, ans = 0, inv = (998244353 + 1) / 2;
int f[2000007];
void insert(int u, int v) {
  other[++cnt] = v, pre[cnt] = last[u], last[u] = cnt;
}
void dfs(int u, int fa) {
  f[u] = 1;
  for (int q = last[u]; q; q = pre[q]) {
    int v = other[q];
    if (v != fa) {
      dfs(v, u);
      int tmp = (998244353 + 1 - (1LL * inv * f[v]) % 998244353) % 998244353;
      f[u] = (1LL * f[u] * tmp) % 998244353;
    }
  }
  ans = (ans + 1 - f[u] + 998244353) % 998244353;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    insert(u, v), insert(v, u);
  }
  dfs(1, -1);
  for (int i = 1; i <= n; i++) ans = (2LL * ans) % 998244353;
  printf("%d\n", (ans + 998244353) % 998244353);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int last[2000007], other[2000007], pre[2000007], cnt = 0;
int n, ans = 0, inv = (998244353 + 1) / 2;
int f[2000007];
void insert(int u, int v) {
  other[++cnt] = v, pre[cnt] = last[u], last[u] = cnt;
}
void dfs(int u, int fa) {
  f[u] = 1;
  for (int q = last[u]; q; q = pre[q]) {
    int v = other[q];
    if (v != fa) {
      dfs(v, u);
      int tmp = (998244353 + 1 - (1LL * inv * f[v]) % 998244353) % 998244353;
      f[u] = (1LL * f[u] * tmp) % 998244353;
    }
  }
  ans = (ans + 1 - f[u] + 998244353) % 998244353;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    insert(u, v), insert(v, u);
  }
  dfs(1, -1);
  for (int i = 1; i <= n; i++) ans = (2LL * ans) % 998244353;
  printf("%d\n", (ans + 998244353) % 998244353);
  return 0;
}

```