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# (Mono-mcs 6.8.0.105) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

namespace ABC149F
{
    class Program
    {
        static void Main(string[] args)
        {
            var N = int.Parse(Console.ReadLine());
            int mod = 1000000007;

            //エッジ一覧
            List<int>[] map = new List<int>[N].Select(x => new List<int>()).ToArray();
            for (int i = 0; i < N - 1; i++)
            {
                var input = ReadInt();
                map[input[0] - 1].Add(input[1] - 1);
                map[input[1] - 1].Add(input[0] - 1);
            }

            //DFS
            firstAccess = new int[N].Select(x => -1).ToArray();
            lastAccess = new int[N];
            DFS(map, 0);

            // 各頂点に関する考察
            long sum = 0;
            for (int i = 0; i < N; i++)
            {
                List<int> counts = new List<int>();

                // 各頂点の行き先
                foreach (var to in map[i])
                {
                    //行き先にある頂点数の算出
                    int count = 0;
                    if (firstAccess[i] < firstAccess[to])
                    {
                        //頂点iより後にアクセスした場合
                        count = (lastAccess[to] - firstAccess[to] + 1) / 2;
                    }
                    else
                    {
                        //頂点iより前の場所の場合
                        count = N - (lastAccess[i] - firstAccess[i] + 1) / 2;
                    }
                    counts.Add(count);
                }

                if (counts.Count > 1)
                {
                    //加算
                    sum += (ModPow(2, N - 1, mod) - counts.Sum(x => ModPow(2, x, mod)) + counts.Count - 1 + mod) % mod;
                    sum %= mod;
                }
            }

            long upper = sum % mod;
            long lower = ModPow(2, N, mod);

            long ans = (upper * ModPow(lower, mod - 2, mod)) % mod;

            Console.WriteLine(ans);
            Console.ReadKey();
        }

        struct Edge
        {
            public int from;
            public int to;
        }

        static int[] firstAccess;
        static int[] lastAccess;
        static int count = 0;
        static void DFS(List<int>[] edges, int edgeNum)
        {
            firstAccess[edgeNum] = count++;

            foreach (var next in edges[edgeNum])
            {
                if (firstAccess[next] != -1) continue;

                DFS(edges, next);
            }

            lastAccess[edgeNum] = count++;
        }



        static int[] ReadInt()
        {
            int[] ret = Console.ReadLine().Split().Select(int.Parse).ToArray();
            return ret;
        }
        static long[] ReadLong()
        {
            long[] ret = Console.ReadLine().Split().Select(long.Parse).ToArray();
            return ret;
        }


        /// <summary>
        /// a^n (mod p)
        /// nが非常に大きい時に有効
        /// </summary>
        /// <param name="a"></param>
        /// <param name="n"></param>
        /// <param name="mod"></param>
        /// <returns></returns>
        public static long ModPow(long a, long n, long p)
        {
            if (n == 0) return 1;
            if (n == 1) return a % p;
            if (n % 2 == 1) return (a * ModPow(a, n - 1, p)) % p;

            long t = ModPow(a, n / 2, p);
            return (t * t) % p;
        }
    }
}