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<iostream>
#include<map>
#include<vector>
#include<cstring>
using namespace std;
const int BUF = 200005;
const int MOD = 1000000007;


int add(int a, int b) {
    return (a + b) % MOD;
}

int sub(int a, int b) {
    return (a + MOD - b) % MOD;
}

int modpow(int p, long long n) {
    if (n == 0) return 1;
    int t = modpow(p, n / 2);
    return n % 2 == 0 ? 1LL * t * t % MOD : 1LL * t * t % MOD * p % MOD;
}


int nNode;
vector<int> adj[BUF];

void read() {
    for (int i = 0; i < BUF; ++i) adj[i].clear();

    cin >> nNode;
    for (int i = 0; i < nNode - 1; ++i) {
        int s, t;
        cin >> s >> t;
        --s, --t;
        adj[s].push_back(t);
        adj[t].push_back(s);
    }
}


int getNumSubtreeNode(int node, int node2nSubtreeNode[BUF], map<pair<int, int>, int> &edge2nSubtreeNode) {
    int &ret = node2nSubtreeNode[node];
    ret = 1;
    
    for (int i = 0; i < adj[node].size(); ++i) {
        int nex = adj[node][i];
        if (node2nSubtreeNode[nex] != -1) continue;
        
        int nexNumSubtreeNode = getNumSubtreeNode(nex, node2nSubtreeNode, edge2nSubtreeNode);
        edge2nSubtreeNode[make_pair(node, nex)] = nexNumSubtreeNode;
        edge2nSubtreeNode[make_pair(nex, node)] = nNode - nexNumSubtreeNode;
        
        ret += nexNumSubtreeNode;
    }
    
    return ret;
}


int rec(int node, int childIdx, int nChose, vector< vector<int> > &dp, map<pair<int, int>, int> &edge2nSubtreeNode) {
    if (childIdx == adj[node].size()) {
        return nChose == 2 ? 1 : 0;
    }

    int &ret = dp[childIdx][nChose];
    if (ret != -1) return ret;
    ret = 0;

    // Choose at least one node from node's childIdx-th subtree
    ret = add(ret, 1LL * rec(node, childIdx + 1, min(nChose + 1, 2), dp, edge2nSubtreeNode) *
              sub(modpow(2, edge2nSubtreeNode[make_pair(node, adj[node][childIdx])]), 1));
    
    // Choose noting
    ret = add(ret, rec(node, childIdx + 1, nChose, dp, edge2nSubtreeNode));
    
    return ret;
}


void work() {
    map<pair<int, int>, int> edge2nSubtreeNode;
    
    int node2nSubtreeNode[BUF];
    memset(node2nSubtreeNode, -1, sizeof(node2nSubtreeNode));
    getNumSubtreeNode(0, node2nSubtreeNode, edge2nSubtreeNode);

    int sum = 0;
    for (int i = 0; i < nNode; ++i) {
        vector<vector<int> > dp(adj[i].size(), vector<int>(3, -1));
        sum = add(sum, rec(i, 0, 0, dp, edge2nSubtreeNode));
    }

    cout << (1LL * sum * modpow(2, 1LL * nNode * (MOD - 2))) % MOD << endl;
}


int main() {
    read();
    work();
    return 0;
}