#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
struct node {
  int next, to;
} e[500020 * 2];
int head[500020], cnt;
long long inv2, f[500020], g[500020];
int n;
inline void adde(int x, int y) {
  e[++cnt].to = y;
  e[cnt].next = head[x];
  head[x] = cnt;
}
inline long long power(long long x, int y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
void dfs(int x, int fa) {
  g[x] = 1;
  for (int i = head[x]; i; i = e[i].next) {
    if (e[i].to == fa) continue;
    dfs(e[i].to, x);
    f[x] = (f[x] + f[e[i].to] + inv2 * g[x] % mod * g[e[i].to]) % mod;
    g[x] = (g[x] - inv2 * g[x] % mod * g[e[i].to] % mod + mod) % mod;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    adde(x, y), adde(y, x);
  }
  inv2 = power(2, mod - 2);
  dfs(1, 0);
  printf("%lld\n", f[1] * power(2, n - 1) * 2 % mod);
}

### 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 long long mod = 998244353;
struct node {
  int next, to;
} e[500020 * 2];
int head[500020], cnt;
long long inv2, f[500020], g[500020];
int n;
inline void adde(int x, int y) {
  e[++cnt].to = y;
  e[cnt].next = head[x];
  head[x] = cnt;
}
inline long long power(long long x, int y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
void dfs(int x, int fa) {
  g[x] = 1;
  for (int i = head[x]; i; i = e[i].next) {
    if (e[i].to == fa) continue;
    dfs(e[i].to, x);
    f[x] = (f[x] + f[e[i].to] + inv2 * g[x] % mod * g[e[i].to]) % mod;
    g[x] = (g[x] - inv2 * g[x] % mod * g[e[i].to] % mod + mod) % mod;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    adde(x, y), adde(y, x);
  }
  inv2 = power(2, mod - 2);
  dfs(1, 0);
  printf("%lld\n", f[1] * power(2, n - 1) * 2 % mod);
}

```