#include <bits/stdc++.h>
using namespace std;
const int Mod = 998244353;
struct Edge {
  int v, nxt;
} e[1000010];
int tot;
int first[500010];
void build(int u, int v) {
  e[++tot] = (Edge){v, first[u]};
  first[u] = tot;
  return;
}
long long f[500010][2];
long long g[500010][2];
long long F[2], G[2];
void dfs(int u, int lst) {
  f[u][0] = 0;
  g[u][0] = 1;
  for (int i = first[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == lst) continue;
    dfs(v, u);
    F[0] = G[0] = F[1] = G[1] = 0;
    G[0] = (G[0] + g[u][0] * g[v][1]) % Mod;
    G[1] = (G[1] + g[u][1] * (g[v][0] + g[v][1]) + g[u][0] * g[v][0]) % Mod;
    F[0] = (F[0] + f[u][0] * g[v][1] + g[u][0] * f[v][1]) % Mod;
    F[1] = (F[1] + f[u][1] * (g[v][0] + g[v][1])) % Mod;
    F[1] = (F[1] + g[u][1] * (f[v][0] + f[v][1])) % Mod;
    F[1] = (F[1] + f[u][0] * g[v][0] + g[u][0] * f[v][0] + g[u][0] * g[v][0]) %
           Mod;
    G[0] = (G[0] + g[u][0] * (g[v][0] + g[v][1])) % Mod;
    G[1] = (G[1] + g[u][1] * (g[v][0] + g[v][1])) % Mod;
    F[0] =
        (F[0] + f[u][0] * (g[v][0] + g[v][1]) + g[u][0] * (f[v][0] + f[v][1])) %
        Mod;
    F[1] =
        (F[1] + f[u][1] * (g[v][0] + g[v][1]) + g[u][1] * (f[v][0] + f[v][1])) %
        Mod;
    f[u][0] = F[0];
    f[u][1] = F[1];
    g[u][0] = G[0];
    g[u][1] = G[1];
  }
  return;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    build(u, v);
    build(v, u);
  }
  dfs(1, 0);
  printf("%lld\n", 2 * (f[1][0] + f[1][1]) % Mod);
  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;
const int Mod = 998244353;
struct Edge {
  int v, nxt;
} e[1000010];
int tot;
int first[500010];
void build(int u, int v) {
  e[++tot] = (Edge){v, first[u]};
  first[u] = tot;
  return;
}
long long f[500010][2];
long long g[500010][2];
long long F[2], G[2];
void dfs(int u, int lst) {
  f[u][0] = 0;
  g[u][0] = 1;
  for (int i = first[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == lst) continue;
    dfs(v, u);
    F[0] = G[0] = F[1] = G[1] = 0;
    G[0] = (G[0] + g[u][0] * g[v][1]) % Mod;
    G[1] = (G[1] + g[u][1] * (g[v][0] + g[v][1]) + g[u][0] * g[v][0]) % Mod;
    F[0] = (F[0] + f[u][0] * g[v][1] + g[u][0] * f[v][1]) % Mod;
    F[1] = (F[1] + f[u][1] * (g[v][0] + g[v][1])) % Mod;
    F[1] = (F[1] + g[u][1] * (f[v][0] + f[v][1])) % Mod;
    F[1] = (F[1] + f[u][0] * g[v][0] + g[u][0] * f[v][0] + g[u][0] * g[v][0]) %
           Mod;
    G[0] = (G[0] + g[u][0] * (g[v][0] + g[v][1])) % Mod;
    G[1] = (G[1] + g[u][1] * (g[v][0] + g[v][1])) % Mod;
    F[0] =
        (F[0] + f[u][0] * (g[v][0] + g[v][1]) + g[u][0] * (f[v][0] + f[v][1])) %
        Mod;
    F[1] =
        (F[1] + f[u][1] * (g[v][0] + g[v][1]) + g[u][1] * (f[v][0] + f[v][1])) %
        Mod;
    f[u][0] = F[0];
    f[u][1] = F[1];
    g[u][0] = G[0];
    g[u][1] = G[1];
  }
  return;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    build(u, v);
    build(v, u);
  }
  dfs(1, 0);
  printf("%lld\n", 2 * (f[1][0] + f[1][1]) % Mod);
  return 0;
}

```