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>
#define ll long long
using namespace std;
const int N = 2e5 + 5, mod = 1e9 + 7;

int sz[N], parent[N], pw[N];
vector<int> adj[N];

int add(int a, int b){
    a = (a + mod) % mod;
    b = (b + mod) % mod;
    return (a + b) % mod;
}
int mul(int a, int b){
    a = (a + mod) % mod;
    b = (b + mod) % mod;
    return 1ll * a * b % mod;
}

int fp(int b, int p){
    if(!p) return 1;
    ll ret = fp(b, p >> 1);
    ret = ret * ret % mod;
    if(p&1) ret = ret * b % mod;
    return ret;

}

void dfs(int node, int par){
    parent[node] = par;
    sz[node] = 1;
    for(auto i: adj[node]){
        if(i == par) continue;
        dfs(i, node);
        sz[node] += sz[i];
    }
}

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    pw[0] = 1;
    for(int i = 1; i < N; i++) pw[i] = pw[i - 1] * 2 % mod;
    int n;
    cin >> n;
    for(int i = 1; i < n; i++){
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    dfs(1, 0);
    int ans = 0;
    for(int i = 1; i <= n; i++){
        ans = add(ans, add(pw[n - 1], -1));
        for(auto j: adj[i]){
            int kam;
            if(j == parent[i]) kam = n - sz[i];
            else kam = sz[j];
            ans = add(ans, -(pw[kam] - 1));
        }
    }
    ans = mul(ans, fp(pw[n], mod - 2));
    cout << ans;

    return 0;

}

