#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5, mod = 998244353, inv2 = (mod + 1) / 2;
int n, x, y, cnt, head[N], f[N], ans;
struct edge {
  int to, nxt;
} e[N << 1];
void adde(int x, int y) {
  e[++cnt].to = y;
  e[cnt].nxt = head[x];
  head[x] = cnt;
}
void dfs(int u, int par) {
  f[u] = 1;
  for (int i = head[u], v; i; i = e[i].nxt)
    if (v = e[i].to, v != par) {
      dfs(v, u);
      f[u] = (long long)f[u] * inv2 % mod * (f[v] + 1) % mod;
    }
  f[u] = (1 + mod - f[u]) % mod;
  ans = (ans + f[u]) % mod;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n - 1); i++)
    scanf("%d%d", &x, &y), adde(x, y), adde(y, x);
  dfs(1, 0);
  for (int i = (1); i <= (n); i++) ans = (ans + ans) % mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 int N = 5e5 + 5, mod = 998244353, inv2 = (mod + 1) / 2;
int n, x, y, cnt, head[N], f[N], ans;
struct edge {
  int to, nxt;
} e[N << 1];
void adde(int x, int y) {
  e[++cnt].to = y;
  e[cnt].nxt = head[x];
  head[x] = cnt;
}
void dfs(int u, int par) {
  f[u] = 1;
  for (int i = head[u], v; i; i = e[i].nxt)
    if (v = e[i].to, v != par) {
      dfs(v, u);
      f[u] = (long long)f[u] * inv2 % mod * (f[v] + 1) % mod;
    }
  f[u] = (1 + mod - f[u]) % mod;
  ans = (ans + f[u]) % mod;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n - 1); i++)
    scanf("%d%d", &x, &y), adde(x, y), adde(y, x);
  dfs(1, 0);
  for (int i = (1); i <= (n); i++) ans = (ans + ans) % mod;
  printf("%d\n", ans);
  return 0;
}

```