Problem:
Score : 600 points

Problem Statement

Given is a tree T with N vertices. The i-th edge connects Vertex A_i and B_i (1 \leq A_i,B_i \leq N).

Now, each vertex is painted black with probability 1/2 and white with probability 1/2, which is chosen independently from other vertices. Then, let S be the smallest subtree (connected subgraph) of T containing all the vertices painted black. (If no vertex is painted black, S is the empty graph.)

Let the holeyness of S be the number of white vertices contained in S. Find the expected holeyness of S.

Since the answer is a rational number, we ask you to print it \bmod 10^9+7, as described in Notes.

Notes

When you print a rational number, first write it as a fraction \frac{y}{x}, where x, y are integers, and x is not divisible by 10^9 + 7
(under the constraints of the problem, such representation is always possible).

Then, you need to print the only integer z between 0 and 10^9 + 6, inclusive, that satisfies xz \equiv y \pmod{10^9 + 7}.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq A_i,B_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
:
A_{N-1} B_{N-1}

Output

Print the expected holeyness of S, \bmod 10^9+7.

Sample Input 1

3
1 2
2 3

Sample Output 1

125000001

If the vertices 1, 2, 3 are painted black, white, black, respectively, the holeyness of S is 1.

Otherwise, the holeyness is 0, so the expected holeyness is 1/8.

Since 8 \times 125000001 \equiv 1 \pmod{10^9+7}, we should print 125000001.

Sample Input 2

4
1 2
2 3
3 4

Sample Output 2

375000003

The expected holeyness is 3/8.

Since 8 \times 375000003 \equiv 3 \pmod{10^9+7}, we should print 375000003.

Sample Input 3

4
1 2
1 3
1 4

Sample Output 3

250000002

The expected holeyness is 1/4.

Sample Input 4

7
4 7
3 1
2 6
5 2
7 1
2 7

Sample Output 4

570312505

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using ll = long long;
using graph=vector<vector<ll>>;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int INF = 1<<30;
const ll mod = 1000000007LL;

ll memo[200050];
bool done[200050];

ll pow2(ll n, ll m){
  if(done[n]) return memo[n];
  done[n]=true;
  ll t = n;
  ll a =2LL;
  ll ret = 1LL;
  for (; n > 0LL; n >>= 1LL, a = a * a % m) {
	if (n % 2LL == 1LL) {
		ret = ret * a % m;
	}
  }
  memo[t]=ret;
  return ret;
}

ll pow(ll a, ll n, ll m) {
	ll ret = 1LL;
	for (; n > 0LL; n >>= 1LL, a = a * a % m) {
		if (n % 2LL == 1LL) {
			ret = ret * a % m;
		}
	}
	return ret;
}

graph children;
vector<bool> seen;
vector<ll> part;
ll N;

void dfs(const graph &g, ll x,ll parent){
  seen[x]=true;
  if(parent!=-1)children[parent].push_back(x);
  for(auto nx:g[x]){
    if(seen[nx]) continue;
    seen[nx]=true;
    dfs(g,nx,x);
  }
  if(parent!=-1) {
    part[parent]=0LL;
    for(auto c:children[parent]) part[parent]+=part[c];
    part[parent]++;
  }
}

ll f(const graph &g,ll i){
  if(children[i].empty()) return 0LL;
  ll ans = pow2(N-1LL,mod)-1LL;
  map<ll,ll> m;
  for(auto x:children[i]){
    m[part[x]]++;
  }
  m[N-part[i]]++;
  for(auto x:m){
    ans=(ans-((pow2(x.first,mod)-1)*x.second)%mod)%mod;
  }
  return ans;
}
int main(){
  cin>>N;
  children.resize(N);
  seen.resize(N);
  part.assign(N,1LL);
  graph g(N);
  for(ll i = 0LL;i < N; i++){
    ll x,y;
    cin>>x>>y;
    g[x-1].push_back(y-1);
    g[y-1].push_back(x-1);
  }
  dfs(g,0,-1);
  ll ans = 0LL;
  for(ll i = 0LL; i < N; i++){
    ans =( ans + f(g,i))%mod;
  }
  if(ans<0) ans+=mod;
  ll y = pow2(N,mod);
  ans = (ans*pow(y,mod-2,mod))%mod;
  cout<<ans<<endl;
}