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)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,s,t) for(int i=s;i<t;i++)
const int mod=1e9+7,inv2=(mod+1)/2;
#define MAXNUM 211111
vector<int> e[MAXNUM];
int tsize[MAXNUM],fa[MAXNUM],p2[MAXNUM];
void dfs(int now,int p)
{
    fa[now]=p,tsize[now]=1;
    for(int k:e[now])if(k!=p)
        dfs(k,now),tsize[now]+=tsize[k];
}
int main()
{
    int n,a,b;scanf("%d",&n);
    p2[0]=1;
    rep(i,1,n+1)p2[i]=p2[i-1]*1ll*inv2%mod;
    rep(i,1,n)scanf("%d%d",&a,&b),e[a].push_back(b),e[b].push_back(a);
    dfs(1,1);int res=0;
    rep(i,1,n+1)
    {
        int num[3]={0};
        num[0]=1;
        for(int k:e[i])
        {
            int x=tsize[k];
            if(k==fa[i])x=n-tsize[i];
            num[2]=(num[2]+num[1]*1ll*(1-p2[x]+mod))%mod;
            num[1]=(num[1]*1ll*p2[x]%mod+num[0]*1ll*(1-p2[x]+mod))%mod;
            num[0]=num[0]*1ll*p2[x]%mod;
        }
        res=(res+num[2]*1ll*inv2%mod)%mod;
    }
    printf("%d\n",res);
}