#include <bits/stdc++.h>
using namespace std;
long long f[500010][2], g[500010][2];
int n, sl, fh;
int t, h[500010];
struct Tre {
  int to, nxt;
} e[500010 << 1];
int rd() {
  sl = 0;
  fh = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') sl = sl * 10 + ch - '0', ch = getchar();
  return sl * fh;
}
void add(int u, int v) {
  e[++t] = (Tre){v, h[u]};
  h[u] = t;
  e[++t] = (Tre){u, h[v]};
  h[v] = t;
}
void dfs(int u, int fa) {
  int v;
  g[u][0] = 1;
  for (int i = h[u]; i; i = e[i].nxt)
    if ((v = e[i].to) != fa) {
      dfs(v, u);
      f[u][1] = (g[u][0] * (f[v][0] + g[v][0]) + g[v][0] * (f[u][0] + g[u][0]) +
                 2 * g[u][1] * (f[v][0] + f[v][1]) +
                 2 * f[u][1] * (g[v][0] + g[v][1])) %
                998244353;
      f[u][0] = (g[u][0] * (f[v][0] + 2 * f[v][1]) +
                 f[u][0] * (g[v][0] + 2 * g[v][1])) %
                998244353;
      g[u][1] =
          (2 * g[u][1] * (g[v][0] + g[v][1]) + g[u][0] * g[v][0]) % 998244353;
      g[u][0] = g[u][0] * (g[v][0] + 2 * g[v][1]) % 998244353;
    }
}
int main() {
  n = rd();
  int x, y;
  for (int i = 1; i < n; ++i) x = rd(), y = rd(), add(x, y);
  dfs(1, 0);
  printf("%I64d\n", (f[1][0] + f[1][1]) % 998244353);
  return 0;
}

### Prompt

Please formulate 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;
long long f[500010][2], g[500010][2];
int n, sl, fh;
int t, h[500010];
struct Tre {
  int to, nxt;
} e[500010 << 1];
int rd() {
  sl = 0;
  fh = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') sl = sl * 10 + ch - '0', ch = getchar();
  return sl * fh;
}
void add(int u, int v) {
  e[++t] = (Tre){v, h[u]};
  h[u] = t;
  e[++t] = (Tre){u, h[v]};
  h[v] = t;
}
void dfs(int u, int fa) {
  int v;
  g[u][0] = 1;
  for (int i = h[u]; i; i = e[i].nxt)
    if ((v = e[i].to) != fa) {
      dfs(v, u);
      f[u][1] = (g[u][0] * (f[v][0] + g[v][0]) + g[v][0] * (f[u][0] + g[u][0]) +
                 2 * g[u][1] * (f[v][0] + f[v][1]) +
                 2 * f[u][1] * (g[v][0] + g[v][1])) %
                998244353;
      f[u][0] = (g[u][0] * (f[v][0] + 2 * f[v][1]) +
                 f[u][0] * (g[v][0] + 2 * g[v][1])) %
                998244353;
      g[u][1] =
          (2 * g[u][1] * (g[v][0] + g[v][1]) + g[u][0] * g[v][0]) % 998244353;
      g[u][0] = g[u][0] * (g[v][0] + 2 * g[v][1]) % 998244353;
    }
}
int main() {
  n = rd();
  int x, y;
  for (int i = 1; i < n; ++i) x = rd(), y = rd(), add(x, y);
  dfs(1, 0);
  printf("%I64d\n", (f[1][0] + f[1][1]) % 998244353);
  return 0;
}

```