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;

#define MOD 1000000007

#define int long long

int n;
vector<int> graph[200010];

int dp1[200010];
int dp2[200010];


int zi[200010];

int pow_mod(int i, int j){
    int ans = 1;
    while(j > 0){
        if(j & 1) ans = (ans * i) % MOD;
        i = (i*i) % MOD;
        j >>= 1;
    }

    return ans;
}


void dp(int idx, int par){
    int ans = 1;
    int ans2 = zi[n-1];

    for(int i = 0;i < graph[idx].size();i++){
        int child = graph[idx][i];
        if(child == par) continue;

        dp(child, idx);

        ans += dp1[child];
        ans2 -= (MOD+zi[dp1[child]]-1)%MOD;
        if(ans2 < 0) ans2 = MOD+ans2;
    }

    dp1[idx] = ans;
    ans2 -= zi[n-ans];
    if(ans2 < 0) ans2 = MOD+ans2;

    dp2[idx] = ans2;
    return;
}

signed main(){
    cin >> n;

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

    zi[0] = 1;
    for(int i = 1;i < 200010;i++){
        zi[i] = zi[i-1] * 2 % MOD;
    }

    dp(0, -1);

    int ans = 0;
    for(int i = 0;i < n;i++){
        ans += dp2[i];
        ans %= MOD;
    }

    cout << ans * pow_mod(zi[n], MOD-2) % MOD << endl;
    return 0;
}

