#include <bits/stdc++.h>
using namespace std;
void Freopen() {
  freopen(
      "title"
      ".in",
      "r", stdin);
  freopen(
      "title"
      ".out",
      "w", stdout);
}
int read() {
  int g = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    g = g * 10 + ch - '0';
    ch = getchar();
  }
  return g * f;
}
const int N = 5e5 + 5;
const int mod = 998244353;
int f[N], n, ans, inv;
vector<int> G[N];
int ksm(int x, int y) {
  int re = 1;
  while (y) {
    if (y & 1) re = 1ll * re * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return re;
}
void dfs(int x, int fa) {
  f[x] = 1;
  for (auto y : G[x])
    if (y ^ fa) {
      dfs(y, x);
      f[x] = 1ll * f[x] * (1 - 1ll * (1 - f[y] + mod) % mod * inv % mod + mod) %
             mod;
    }
  f[x] = (1 - f[x] + mod) % mod;
  ans = (ans + 2 * f[x] % mod) % mod;
}
signed main() {
  inv = ksm(2, mod - 2);
  n = read();
  for (int i = (1); i <= (n - 1); i++) {
    int x = read(), y = read();
    G[x].push_back(y);
    G[y].push_back(x);
  }
  dfs(1, 0);
  return cout << 1ll * ans * ksm(2, n - 1) % mod, signed();
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
void Freopen() {
  freopen(
      "title"
      ".in",
      "r", stdin);
  freopen(
      "title"
      ".out",
      "w", stdout);
}
int read() {
  int g = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    g = g * 10 + ch - '0';
    ch = getchar();
  }
  return g * f;
}
const int N = 5e5 + 5;
const int mod = 998244353;
int f[N], n, ans, inv;
vector<int> G[N];
int ksm(int x, int y) {
  int re = 1;
  while (y) {
    if (y & 1) re = 1ll * re * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return re;
}
void dfs(int x, int fa) {
  f[x] = 1;
  for (auto y : G[x])
    if (y ^ fa) {
      dfs(y, x);
      f[x] = 1ll * f[x] * (1 - 1ll * (1 - f[y] + mod) % mod * inv % mod + mod) %
             mod;
    }
  f[x] = (1 - f[x] + mod) % mod;
  ans = (ans + 2 * f[x] % mod) % mod;
}
signed main() {
  inv = ksm(2, mod - 2);
  n = read();
  for (int i = (1); i <= (n - 1); i++) {
    int x = read(), y = read();
    G[x].push_back(y);
    G[y].push_back(x);
  }
  dfs(1, 0);
  return cout << 1ll * ans * ksm(2, n - 1) % mod, signed();
}

```