#include <bits/stdc++.h>
const int maxn = 1111111, mod = 998244353;
int n;
struct Edge {
  int to;
  Edge *ne;
} el[maxn], *la[maxn], *cl = el;
int f[maxn];
int ans;
int In() {
  register int c, f, s;
  for (; c = getchar(), c != '-' && !isdigit(c);)
    ;
  for (s = (f = c == '-') ? 0 : c - '0'; c = getchar(), isdigit(c);)
    s = s * 10 + c - '0';
  return f ? -s : s;
}
void ins(register int u, register int v) {
  *++cl = (Edge){v, la[u]}, la[u] = cl;
  *++cl = (Edge){u, la[v]}, la[v] = cl;
}
void dfs(register int x, register int y) {
  for (register Edge *i = la[x]; i; i = i->ne)
    if (i->to != y) dfs(i->to, x);
  for (register Edge *i = la[x]; i; i = i->ne)
    if (i->to != y)
      f[x] =
          (f[x] + 499122177ll * (mod + 1 - f[x]) % mod * (mod + 1 - f[i->to])) %
          mod;
  ans = (ans + f[x]) % mod;
}
int main() {
  n = In();
  for (register int i = 1, u, v; i < n; ++i) u = In(), v = In(), ins(u, v);
  dfs(1, 0);
  for (register int i = 1; i <= n; ++i) ans = 2ll * ans % mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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>
const int maxn = 1111111, mod = 998244353;
int n;
struct Edge {
  int to;
  Edge *ne;
} el[maxn], *la[maxn], *cl = el;
int f[maxn];
int ans;
int In() {
  register int c, f, s;
  for (; c = getchar(), c != '-' && !isdigit(c);)
    ;
  for (s = (f = c == '-') ? 0 : c - '0'; c = getchar(), isdigit(c);)
    s = s * 10 + c - '0';
  return f ? -s : s;
}
void ins(register int u, register int v) {
  *++cl = (Edge){v, la[u]}, la[u] = cl;
  *++cl = (Edge){u, la[v]}, la[v] = cl;
}
void dfs(register int x, register int y) {
  for (register Edge *i = la[x]; i; i = i->ne)
    if (i->to != y) dfs(i->to, x);
  for (register Edge *i = la[x]; i; i = i->ne)
    if (i->to != y)
      f[x] =
          (f[x] + 499122177ll * (mod + 1 - f[x]) % mod * (mod + 1 - f[i->to])) %
          mod;
  ans = (ans + f[x]) % mod;
}
int main() {
  n = In();
  for (register int i = 1, u, v; i < n; ++i) u = In(), v = In(), ins(u, v);
  dfs(1, 0);
  for (register int i = 1; i <= n; ++i) ans = 2ll * ans % mod;
  printf("%d\n", ans);
  return 0;
}

```