#include <bits/stdc++.h>
using namespace std;
struct edge {
  int to, lt;
} e[500010 << 1];
int in[500010], cnt;
void add(int x, int y) {
  e[++cnt].to = y;
  e[cnt].lt = in[x];
  in[x] = cnt;
}
void addedge(int x, int y) {
  add(x, y);
  add(y, x);
}
int ksm(int bs, int mi) {
  int ans = 1;
  while (mi) {
    if (mi & 1) ans = (long long)ans * bs % 998244353;
    bs = (long long)bs * bs % 998244353;
    mi >>= 1;
  }
  return ans;
}
int f[500010], d[500010], inv = ksm(2, 998244353 - 2), n;
int tot;
void dfs(int x, int fa) {
  if (d[x] == 1 && x != fa) return;
  int tmp = 1;
  for (int i = in[x]; i; i = e[i].lt) {
    int y = e[i].to;
    if (fa == y) continue;
    dfs(y, x);
    tmp = (tmp * (1ll + f[y]) % 998244353) * inv % 998244353;
  }
  f[x] = 1 - tmp + 998244353 >= 998244353 ? 1 - tmp + 998244353 - 998244353
                                          : 1 - tmp + 998244353;
  tot = tot + f[x] >= 998244353 ? tot + f[x] - 998244353 : tot + f[x];
}
int main() {
  int x, y;
  scanf("%d", &n);
  for (int i = 1; i < n; i++)
    scanf("%d%d", &x, &y), d[x]++, d[y]++, addedge(x, y);
  dfs(1, 1);
  printf("%d\n", (long long)tot * ksm(2, n) % 998244353);
  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;
struct edge {
  int to, lt;
} e[500010 << 1];
int in[500010], cnt;
void add(int x, int y) {
  e[++cnt].to = y;
  e[cnt].lt = in[x];
  in[x] = cnt;
}
void addedge(int x, int y) {
  add(x, y);
  add(y, x);
}
int ksm(int bs, int mi) {
  int ans = 1;
  while (mi) {
    if (mi & 1) ans = (long long)ans * bs % 998244353;
    bs = (long long)bs * bs % 998244353;
    mi >>= 1;
  }
  return ans;
}
int f[500010], d[500010], inv = ksm(2, 998244353 - 2), n;
int tot;
void dfs(int x, int fa) {
  if (d[x] == 1 && x != fa) return;
  int tmp = 1;
  for (int i = in[x]; i; i = e[i].lt) {
    int y = e[i].to;
    if (fa == y) continue;
    dfs(y, x);
    tmp = (tmp * (1ll + f[y]) % 998244353) * inv % 998244353;
  }
  f[x] = 1 - tmp + 998244353 >= 998244353 ? 1 - tmp + 998244353 - 998244353
                                          : 1 - tmp + 998244353;
  tot = tot + f[x] >= 998244353 ? tot + f[x] - 998244353 : tot + f[x];
}
int main() {
  int x, y;
  scanf("%d", &n);
  for (int i = 1; i < n; i++)
    scanf("%d%d", &x, &y), d[x]++, d[y]++, addedge(x, y);
  dfs(1, 1);
  printf("%d\n", (long long)tot * ksm(2, n) % 998244353);
  return 0;
}

```