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: Accepted)


// F - Surrounded Nodes

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
// const int INF = 2147483647;
// const ll INF = 9223372036854775807;
const ll MOD = 1e9 + 7;

struct Node {
	vector<int> neighbors;
};

int N;
Node nodes[200000];
vector<int> node_cnt[200000]; // node_cnt[i][j]: nodes[i]のj番目のneighborから先のノードの数

// mod p （pは素数）の下でのべき乗を求める
ll mod_pow(ll a, ll n, ll p) {
		ll result = 1;
		while (n > 0) {
			if (n & 1) result = result * a % p;
			a = a * a % p;
			n >>= 1;
		}
		return result;
	};

// mod p （pは素数）の下での逆元を求める（フェルマーの小定理を利用）
ll mod_inverse(ll a, ll p) {
	return mod_pow(a, p-2, p);
}

// mod p （pは素数）の下での除算 (a / b)
ll mod_devide(ll a, ll b, ll p) {
	return a * mod_inverse(b, p) % p;
}

// nodes[i] から先（nodes[i]を含む）のノードの数を返す
// 同時にprev側のノードの数も計算
int dfs(int i, int prev_i) {
	int result = 1;
	int prev_j = -1;
	for (int j=0; j<nodes[i].neighbors.size(); j++) {
		int next_i = nodes[i].neighbors[j];

		if (next_i == prev_i) {
			prev_j = j;
			continue;
		}

		if (node_cnt[i][j] != -1) {
			result += node_cnt[i][j];
			continue;
		}

		node_cnt[i][j] = dfs(next_i, i);
		result += node_cnt[i][j];
	};

	if (prev_j != -1) {
		node_cnt[i][prev_j] = N - result;
	}

	return result;
}

int main() {
	cin >> N;

	int A, B;
	for (int i=0; i<N-1; i++) {
		cin >> A >> B;
		A--;
		B--;
		nodes[A].neighbors.push_back(B);
		node_cnt[A].push_back(-1);
		nodes[B].neighbors.push_back(A);
		node_cnt[B].push_back(-1);
	}

	// node_cnt[i][j]を求める
	dfs(0, -1);

	/*
	// **** debug ****
	for (int i=0; i<N; i++) {
		cout << "node " << i + 1 << " :";
		for (int n : node_cnt[i]) {
			cout << " " << n;
		}
		cout << endl;
	}
	*/

	// nodeごとに穴あき度の期待値を算出し、それを合計する
	// （y/xについて、xの逆元にyを掛けたものがz）
	ll ans = 0;
	ll inv2 = mod_inverse(2, MOD); // 2の逆元

	// mod_powを予め計算
	ll prob_all_white = mod_pow(inv2, N-1, MOD);

	/*
	unordered_map<ll, ll> mod_pow_inv2;
	for (int i=0; i<N; i++) {
		for (int j=0; j<nodes[i].neighbors.size(); j++) {
			if (mod_pow_inv2.find(node_cnt[i][j]) == mod_pow_inv2.end()) {
				mod_pow_inv2[node_cnt[i][j]] = mod_pow(inv2, node_cnt[i][j], MOD);
			}
		}
	}
	*/

	/*
	for (int i=0; i<N; i++) {
		if (nodes[i].neighbors.size() == 1) continue; // 端点は期待値0

		// nodes[i]から出ている部分木のうち、nodes[i].neighbors[j] からの木だけが黒を含む確率
		ll prob_one_black = 0;
		for (int j=0; j<nodes[i].neighbors.size(); j++) {
			ll prob = (1 - mod_pow_inv2[node_cnt[i][j]] + MOD) % MOD; // 黒を含む部分木
			prob = (prob * mod_pow(inv2, N - node_cnt[i][j] - 1, MOD)) % MOD; // それ以外
			prob_one_black = (prob_one_black + prob) % MOD;
		}

		ll prob = (1 - prob_all_white + MOD) % MOD;
		prob = (prob - prob_one_black + MOD) % MOD;
		prob = (prob * inv2) % MOD; // そもそもnodes[i]が白

		// 確率がそのまま穴あき度の期待値になる
		ans = (ans + prob) % MOD;
	}
	*/
	for (int i=0; i<N; i++) {
		if (nodes[i].neighbors.size() == 1) continue; // 端点は期待値0

		// nodes[i]から出ている部分木のうち、nodes[i].neighbors[j] からの木だけが黒を含む確率
		ll prob_one_black = 0;
		for (int j=0; j<nodes[i].neighbors.size(); j++) {
			ll prob = (1 - mod_pow(inv2, node_cnt[i][j], MOD) + MOD) % MOD; // 黒を含む部分木
			prob = (prob * mod_pow(inv2, N - node_cnt[i][j] - 1, MOD)) % MOD; // それ以外
			prob_one_black = (prob_one_black + prob) % MOD;
		}

		ll prob = (1 - prob_all_white + MOD) % MOD;
		prob = (prob - prob_one_black + MOD) % MOD;
		prob = (prob * inv2) % MOD; // そもそもnodes[i]が白

		// 確率がそのまま穴あき度の期待値になる
		ans = (ans + prob) % MOD;
	}

	cout << ans << endl;

	return 0;
}