#include <bits/stdc++.h>
const int MAXN = 500010;
const int Dollar1 = 998244353;
const int INV2 = (Dollar1 + 1) / 2;
inline int up(int x) { return x -= Dollar1, x + (x >> 31 & Dollar1); }
inline void mulr(int& x, const int y) {
  x = static_cast<long long>(x) * y % Dollar1;
}
inline int mul(const int x, const int y) {
  return static_cast<long long>(x) * y % Dollar1;
}
int head[MAXN], nxt[MAXN << 1], to[MAXN << 1], tot;
inline void addedge(int b, int e) {
  nxt[++tot] = head[b];
  to[head[b] = tot] = e;
  nxt[++tot] = head[e];
  to[head[e] = tot] = b;
}
int f[MAXN];
void dfs(int u, int fa) {
  int prod = 1;
  for (int i = head[u]; i; i = nxt[i])
    if (to[i] != fa) {
      dfs(to[i], u);
      mulr(prod, mul(up(f[to[i]] + 1), INV2));
    }
  f[u] = up(Dollar1 + 1 - prod);
}
int n, t1, t2;
int main() {
  scanf("%d", &n);
  for (int i = 1; i != n; ++i) {
    scanf("%d%d", &t1, &t2);
    addedge(t1, t2);
  }
  dfs(1, 0);
  int ans = 0;
  for (int i = 1; i <= n; ++i) ans = up(ans + f[i]);
  for (int i = 1; i <= n; ++i) mulr(ans, 2);
  printf("%d\n", ans);
  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>
const int MAXN = 500010;
const int Dollar1 = 998244353;
const int INV2 = (Dollar1 + 1) / 2;
inline int up(int x) { return x -= Dollar1, x + (x >> 31 & Dollar1); }
inline void mulr(int& x, const int y) {
  x = static_cast<long long>(x) * y % Dollar1;
}
inline int mul(const int x, const int y) {
  return static_cast<long long>(x) * y % Dollar1;
}
int head[MAXN], nxt[MAXN << 1], to[MAXN << 1], tot;
inline void addedge(int b, int e) {
  nxt[++tot] = head[b];
  to[head[b] = tot] = e;
  nxt[++tot] = head[e];
  to[head[e] = tot] = b;
}
int f[MAXN];
void dfs(int u, int fa) {
  int prod = 1;
  for (int i = head[u]; i; i = nxt[i])
    if (to[i] != fa) {
      dfs(to[i], u);
      mulr(prod, mul(up(f[to[i]] + 1), INV2));
    }
  f[u] = up(Dollar1 + 1 - prod);
}
int n, t1, t2;
int main() {
  scanf("%d", &n);
  for (int i = 1; i != n; ++i) {
    scanf("%d%d", &t1, &t2);
    addedge(t1, t2);
  }
  dfs(1, 0);
  int ans = 0;
  for (int i = 1; i <= n; ++i) ans = up(ans + f[i]);
  for (int i = 1; i <= n; ++i) mulr(ans, 2);
  printf("%d\n", ans);
  return 0;
}

```