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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;

public class Main {
	static int mod = (int) 1e9 + 7;
	static List<ArrayList<Integer>> to;
	static int dp[];
	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		int N = fs.nextInt();
		to = new ArrayList<>();
		for(int i=0;i<N;i++) to.add(new ArrayList<Integer>());
		int edges[][] = new int[N-1][2]; //片方の頂点だけ分かればいい。ただしdfsの原点はだめ。
		for(int i=0;i<N-1;i++) {
			int A = fs.nextInt()-1, B = fs.nextInt()-1;
			to.get(A).add(B);
			to.get(B).add(A);
			edges[i][0] = B;
			edges[i][1] = A;
		}
		dp = new int[N];
		rec(0, -1);
		long mulsum = 0;
		for(int e[]:edges) {
			int candidate1 = e[0];
			int candidate2 = e[1];
			int child = candidate1;
			if(dp[candidate1]>dp[candidate2])child = candidate2;
			
			long a = dp[child];
			long b = N - a;

			long mul = (modpow(2, a) - 1)*(modpow(2, b) - 1)%mod;			
			mulsum += mul;
			mulsum %= mod;
			//分母は全てmodpow(2, n)になる
		}
		long cases = modpow(2, N);
		long PN = ((mulsum + (cases - (N+1) + mod)%mod ) %mod + N)%mod;//全ケースの含まれる頂点の合計
		long BN = N * modpow(2, N-1) %mod; 
		long WN = (PN - BN + mod)%mod;//期待値を出すまではできたようだけど
		long ans = WN * modpow(cases, mod-2);
		ans %= mod;
		
		System.out.println(ans);
	}
	
	static int rec(int v, int parent) {
		int res = 1;
		for(int next:to.get(v)) {
			if(next == parent) continue;
			res += rec(next, v);
		}
		return dp[v] = res;
	}
	

	static long modpow(long x, long y) {
		if(y==0) return 1;
		if(y%2!=0) return x * modpow(x, y-1) % mod;
		long tmp = modpow(x, y/2);
		return tmp * tmp % mod; 
	}
}

//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}