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.

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.

Examples

Input

3
1 2
2 3


Output

125000001


Input

4
1 2
2 3
3 4


Output

375000003


Input

4
1 2
1 3
1 4


Output

250000002


Input

7
4 7
3 1
2 6
5 2
7 1
2 7


Output

570312505
#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;

int n;
vector<vector<int>> graph;
vector<long long int> pow2;

long long int ans = 0;
int DFS(int v, int parent_v = -1){
    vector<int> subtree_sizes;
    int total = 1;// self

    for(int child_v : graph[v]){
        if(child_v == parent_v) continue;

        int sub = DFS(child_v, v);
        subtree_sizes.push_back(sub);
        total += sub;
    }
    if(parent_v != -1) subtree_sizes.push_back(n - total);

    long long int add = pow2[n-1] - 1;
    for(int s : subtree_sizes){
        add -= pow2[s] - 1;
        add = (add % MOD + MOD) % MOD;
    }
    ans += add;
    ans %= MOD;

    return total;
}

long long int modpow(long long int b, long long int p){
    long long int num = 1, power = b;
    for(int i=0; i<=30; i++){
        if((p>>i) & 1){
            num *= power;
            num %= MOD;
        }

        power *= power;
        power %= MOD;
    }
    return num;
}

int main(){
    cin >> n;

    graph.resize(n);
    for(int i=0; i<n-1; i++){
        int a, b;
        cin >> a >> b;
        a -= 1;
        b -= 1;
        graph[a].push_back(b);
        graph[b].push_back(a);
    }

    pow2.resize(n+1, 1);// pow2[i] = 2^i
    for(int i=0; i<n; i++){
        pow2[i+1] = pow2[i] * 2;
        pow2[i+1] %= MOD;
    }

    DFS(0);
    //cout << ans << " " << pow2[n] << " ";

    long long int x = modpow(pow2[n], MOD-2);
    //cout << ((x * pow2[n]) % MOD)<< endl;
    ans *= x;
    ans %= MOD;
    cout << ans << endl;
    return 0;
}