#include <bits/stdc++.h>
using namespace std;
namespace fast_io {
char buf[1 << 12], *p1 = buf, *p2 = buf, sr[1 << 23], z[23], nc;
int C = -1, Z = 0, Bi = 0, ny;
inline char gc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 12, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  ny = 1;
  while (nc = gc(), (nc < 48 || nc > 57) && nc != EOF)
    if (nc == 45) ny = -1;
  Bi = 1;
  if (nc < 0) return nc;
  x = nc - 48;
  while (nc = gc(), 47 < nc && nc < 58 && nc != EOF)
    x = (x << 3) + (x << 1) + (nc ^ 48), Bi++;
  return x * ny;
}
inline double find() {
  int a = read(), y = ny, b = (nc != '.') ? 0 : read();
  return (b ? a + (double)b / pow(10, Bi) * y : a);
}
inline int gs(char *s) {
  char c, *t = s;
  while (c = gc(), c <= 32)
    ;
  *s++ = c;
  while (c = gc(), c > 32) *s++ = c;
  return s - t;
}
inline void ot() {
  fwrite(sr, 1, C + 1, stdout);
  C = -1;
}
inline void flush() {
  if (C > 1 << 22) ot();
}
template <typename T>
inline void write(T x, char t) {
  int y = 0;
  if (x < 0) y = 1, x = -x;
  while (z[++Z] = x % 10 + 48, x /= 10)
    ;
  if (y) z[++Z] = '-';
  while (sr[++C] = z[Z], --Z)
    ;
  sr[++C] = t;
  flush();
}
inline void write(char *s) {
  int l = strlen(s);
  for (int i = 0; i < l; i++) sr[++C] = *s++;
  sr[++C] = '\n';
  flush();
}
}  // namespace fast_io
using namespace fast_io;
const int N = 5e5 + 5, p = 998244353, iv2 = 499122177;
int n, c, ans, f[N];
vector<int> e[N];
inline void add(int u, int v) { e[u].push_back(v), e[v].push_back(u); }
inline void M(int &x) {
  x -= p;
  x += x >> 31 & p;
}
void dfs(int u, int fa) {
  int res = 1;
  for (int v : e[u])
    if (v ^ fa) dfs(v, u), res = 1ll * res * (f[v] + 1) % p * iv2 % p;
  M(ans += (f[u] = p + 1 - res));
}
int main() {
  n = read(), c = 1;
  for (int i = 1; i < n; i++) add(read(), read()), M(c <<= 1);
  dfs(1, 0);
  write(2ll * ans * c % p, '\n');
  return ot(), 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
namespace fast_io {
char buf[1 << 12], *p1 = buf, *p2 = buf, sr[1 << 23], z[23], nc;
int C = -1, Z = 0, Bi = 0, ny;
inline char gc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 12, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  ny = 1;
  while (nc = gc(), (nc < 48 || nc > 57) && nc != EOF)
    if (nc == 45) ny = -1;
  Bi = 1;
  if (nc < 0) return nc;
  x = nc - 48;
  while (nc = gc(), 47 < nc && nc < 58 && nc != EOF)
    x = (x << 3) + (x << 1) + (nc ^ 48), Bi++;
  return x * ny;
}
inline double find() {
  int a = read(), y = ny, b = (nc != '.') ? 0 : read();
  return (b ? a + (double)b / pow(10, Bi) * y : a);
}
inline int gs(char *s) {
  char c, *t = s;
  while (c = gc(), c <= 32)
    ;
  *s++ = c;
  while (c = gc(), c > 32) *s++ = c;
  return s - t;
}
inline void ot() {
  fwrite(sr, 1, C + 1, stdout);
  C = -1;
}
inline void flush() {
  if (C > 1 << 22) ot();
}
template <typename T>
inline void write(T x, char t) {
  int y = 0;
  if (x < 0) y = 1, x = -x;
  while (z[++Z] = x % 10 + 48, x /= 10)
    ;
  if (y) z[++Z] = '-';
  while (sr[++C] = z[Z], --Z)
    ;
  sr[++C] = t;
  flush();
}
inline void write(char *s) {
  int l = strlen(s);
  for (int i = 0; i < l; i++) sr[++C] = *s++;
  sr[++C] = '\n';
  flush();
}
}  // namespace fast_io
using namespace fast_io;
const int N = 5e5 + 5, p = 998244353, iv2 = 499122177;
int n, c, ans, f[N];
vector<int> e[N];
inline void add(int u, int v) { e[u].push_back(v), e[v].push_back(u); }
inline void M(int &x) {
  x -= p;
  x += x >> 31 & p;
}
void dfs(int u, int fa) {
  int res = 1;
  for (int v : e[u])
    if (v ^ fa) dfs(v, u), res = 1ll * res * (f[v] + 1) % p * iv2 % p;
  M(ans += (f[u] = p + 1 - res));
}
int main() {
  n = read(), c = 1;
  for (int i = 1; i < n; i++) add(read(), read()), M(c <<= 1);
  dfs(1, 0);
  write(2ll * ans * c % p, '\n');
  return ot(), 0;
}

```