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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) begin(v),end(v)
#define fi first
#define se second
template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }
using ll = long long;
using pii = pair<int, int>;
constexpr ll INF = 1ll<<30;
constexpr ll longINF = 1ll<<60;
constexpr ll MOD = 1000000007;
constexpr bool debug = 0;
//---------------------------------//

ll mod_pow(ll x, ll n, ll mod = MOD) {
	if (n == 0) return 1;
	ll res = mod_pow(x * x % mod, n / 2, mod);
	if (n & 1) res = res * x % mod;
	
	return res;
}

int N;
vector<int> g[212345];
int cnt[212345], oya[212345];

int dfs(int u, int par) {
	int res = 1;
	for (int v : g[u]) {
		if (v == par) continue;
		oya[v] = u;
		res += dfs(v, u);
	}
	return cnt[u] = res;
}

int main() {
	cin >> N;
	REP(i, N - 1) {
		int a, b;
		scanf("%d %d", &a, &b);
		--a; --b;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	
	if (N == 2) {
		cout << 0 << endl;
		return 0;
	}
	
	dfs(0, -1);
	
	ll ans = 0;
	REP(i, N) {
		// ll cur = (cnt[i] - 1) * (N - cnt[i]) % MOD;
		// cur = cur * mod_pow(2, N - 3) % MOD;
		
		ll cur = mod_pow(mod_pow(2, N, MOD), MOD - 2, MOD);
		vector<int> childs;
		if (N - cnt[i] > 0) childs.push_back(N - cnt[i]);
		for (int v : g[i]) if (v != oya[i]) {
			childs.push_back(cnt[v]);
		}
		if (childs.size() < 2) continue;
		vector<vector<ll> > dp(childs.size() + 1, vector<ll>(3, 0));
		dp[0][0] = 1;
		REP(i, childs.size()) REP(j, 3) {
			if (j < 2) {
				ll pat = (mod_pow(2, childs[i], MOD) + MOD) % MOD;
				pat = (pat - 1 + MOD) % MOD;
				(dp[i + 1][j + 1] += dp[i][j] * pat % MOD) %= MOD;
			}
			else {
				ll pat = (mod_pow(2, childs[i], MOD) - childs[i] + MOD) % MOD;
				pat = (pat - 1 + MOD) % MOD;
				(dp[i + 1][j] += dp[i][j] * pat) %= MOD;
				
			}
		}
		cur = cur * dp[childs.size()][2] % MOD;
		ans = (ans + cur) % MOD;
	}
	cout << ans << endl;
	return 0;
}