#include <bits/stdc++.h>
using namespace std;
const int Mod = 998244353, MAXN = 5e5 + 5;
char buf[1 << 12], *pp1 = buf, *pp2 = buf, nc;
int ny;
inline char gc() {
  return pp1 == pp2 &&
                 (pp2 = (pp1 = buf) + fread(buf, 1, 1 << 12, stdin), pp1 == pp2)
             ? EOF
             : *pp1++;
}
inline int read() {
  int x = 0;
  for (ny = 1; nc = gc(), (nc < 48 || nc > 57) && nc != EOF;)
    if (nc == 45) ny = -1;
  if (nc < 0) return nc;
  for (x = nc - 48; nc = gc(), 47 < nc && nc < 58 && nc != EOF;
       x = (x << 3) + (x << 1) + (nc ^ 48))
    ;
  return x * ny;
}
int n, f[MAXN][2][2];
vector<int> G[MAXN];
void dfs(int x, int fa) {
  f[x][0][0] = 1;
  for (auto y : G[x])
    if (y != fa) {
      dfs(y, x);
      int t1 = f[x][0][0], t2 = f[x][0][1], t3 = f[x][1][0], t4 = f[x][1][1];
      f[x][0][0] = 1ll * t1 * (0ll + f[y][0][0] + 2 * f[y][1][0]) % Mod;
      f[x][1][0] = (1ll * t1 * f[y][0][0] % Mod +
                    2ll * t3 * (f[y][0][0] + f[y][1][0]) % Mod) %
                   Mod;
      f[x][0][1] = (1ll * t2 * (0ll + f[y][0][0] + 2 * f[y][1][0]) % Mod +
                    1ll * t1 * (0ll + f[y][0][1] + 2 * f[y][1][1]) % Mod) %
                   Mod;
      f[x][1][1] = (2ll * t4 * (f[y][0][0] + f[y][1][0]) % Mod +
                    1ll * t2 * f[y][0][0] % Mod +
                    1ll * t1 * (f[y][0][0] + f[y][0][1]) % Mod +
                    2ll * t3 * (f[y][0][1] + f[y][1][1]) % Mod) %
                   Mod;
    }
}
int main() {
  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), cout << (f[1][0][1] + f[1][1][1]) % Mod * 2 % Mod << "\n";
  return 0;
}

### Prompt

Please create a solution in Cpp 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 int Mod = 998244353, MAXN = 5e5 + 5;
char buf[1 << 12], *pp1 = buf, *pp2 = buf, nc;
int ny;
inline char gc() {
  return pp1 == pp2 &&
                 (pp2 = (pp1 = buf) + fread(buf, 1, 1 << 12, stdin), pp1 == pp2)
             ? EOF
             : *pp1++;
}
inline int read() {
  int x = 0;
  for (ny = 1; nc = gc(), (nc < 48 || nc > 57) && nc != EOF;)
    if (nc == 45) ny = -1;
  if (nc < 0) return nc;
  for (x = nc - 48; nc = gc(), 47 < nc && nc < 58 && nc != EOF;
       x = (x << 3) + (x << 1) + (nc ^ 48))
    ;
  return x * ny;
}
int n, f[MAXN][2][2];
vector<int> G[MAXN];
void dfs(int x, int fa) {
  f[x][0][0] = 1;
  for (auto y : G[x])
    if (y != fa) {
      dfs(y, x);
      int t1 = f[x][0][0], t2 = f[x][0][1], t3 = f[x][1][0], t4 = f[x][1][1];
      f[x][0][0] = 1ll * t1 * (0ll + f[y][0][0] + 2 * f[y][1][0]) % Mod;
      f[x][1][0] = (1ll * t1 * f[y][0][0] % Mod +
                    2ll * t3 * (f[y][0][0] + f[y][1][0]) % Mod) %
                   Mod;
      f[x][0][1] = (1ll * t2 * (0ll + f[y][0][0] + 2 * f[y][1][0]) % Mod +
                    1ll * t1 * (0ll + f[y][0][1] + 2 * f[y][1][1]) % Mod) %
                   Mod;
      f[x][1][1] = (2ll * t4 * (f[y][0][0] + f[y][1][0]) % Mod +
                    1ll * t2 * f[y][0][0] % Mod +
                    1ll * t1 * (f[y][0][0] + f[y][0][1]) % Mod +
                    2ll * t3 * (f[y][0][1] + f[y][1][1]) % Mod) %
                   Mod;
    }
}
int main() {
  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), cout << (f[1][0][1] + f[1][1][1]) % Mod * 2 % Mod << "\n";
  return 0;
}

```