#include <bits/stdc++.h>
using namespace std;
int sz[500005], pw[500005];
long long ans[500005];
struct edge {
  int u, v;
  edge *next;
} p[500005 << 1], *h[500005];
int egcnt = 0;
void addedge(int u, int v) {
  edge *n1 = &p[egcnt++];
  n1->u = u, n1->v = v;
  n1->next = h[u], h[u] = n1;
}
void dfs(int a, int fa) {
  sz[a] = 1;
  long long nans = 1;
  for (edge *p = h[a]; p; p = p->next) {
    if (p->v == fa) continue;
    dfs(p->v, a);
    sz[a] += sz[p->v];
    nans *= (pw[sz[p->v] - 1] + ans[p->v]), nans %= 998244353;
  }
  ans[a] = pw[sz[a] - 1] - nans;
  ans[a] %= 998244353;
}
int main() {
  pw[0] = 1;
  for (int i = 1; i < 500005; i++) pw[i] = pw[i - 1] * 2 % 998244353;
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v), addedge(v, u);
  }
  dfs(1, 0);
  long long res = 0;
  for (int i = 1; i <= n; i++) ans[i] *= pw[n - sz[i]], ans[i] %= 998244353;
  for (int i = 1; i <= n; i++) res += ans[i];
  res *= 2, res %= 998244353;
  if (res < 0) res += 998244353;
  printf("%lld\n", res);
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int sz[500005], pw[500005];
long long ans[500005];
struct edge {
  int u, v;
  edge *next;
} p[500005 << 1], *h[500005];
int egcnt = 0;
void addedge(int u, int v) {
  edge *n1 = &p[egcnt++];
  n1->u = u, n1->v = v;
  n1->next = h[u], h[u] = n1;
}
void dfs(int a, int fa) {
  sz[a] = 1;
  long long nans = 1;
  for (edge *p = h[a]; p; p = p->next) {
    if (p->v == fa) continue;
    dfs(p->v, a);
    sz[a] += sz[p->v];
    nans *= (pw[sz[p->v] - 1] + ans[p->v]), nans %= 998244353;
  }
  ans[a] = pw[sz[a] - 1] - nans;
  ans[a] %= 998244353;
}
int main() {
  pw[0] = 1;
  for (int i = 1; i < 500005; i++) pw[i] = pw[i - 1] * 2 % 998244353;
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v), addedge(v, u);
  }
  dfs(1, 0);
  long long res = 0;
  for (int i = 1; i <= n; i++) ans[i] *= pw[n - sz[i]], ans[i] %= 998244353;
  for (int i = 1; i <= n; i++) res += ans[i];
  res *= 2, res %= 998244353;
  if (res < 0) res += 998244353;
  printf("%lld\n", res);
  return 0;
}

```