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;
typedef long long ll;
#define mp make_pair
#define pb push_back
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)

const ll mod=1e9+7;
ll mu(ll a,ll b){
  return (ll)a*b%mod;
}
ll ad(ll a,ll b){
  if((a+=b)>=mod)a-=mod;
  return a;
}
ll pw(ll a,ll b,ll c=1LL){
  for(;b;b>>=1,a=mu(a,a))if(b&1)c=mu(c,a);
  return c;
}
ll dv(ll a,ll b){
  return mu(a,pw(b,mod-2));
}

ll ans=0;
const int n_max=2e5+10;
vector<int> g[n_max];
ll pwall;
ll n;


int dfs(int v, int p){
  vector<int> kv;
  int sum=1;
  for(auto i:g[v]){
    if(i==p) continue;
    int k=dfs(i,v);
    sum+=k;
    kv.pb(k);
  }
  kv.pb(n-sum);
  ll out=1;
  rep(j,kv.size()){
    //    cout << kv[j] << endl;
    out = ad(out,ad(pw(2,kv[j])-1+mod,0));    
  }
  ll ok = ad(pwall - out +mod,0);
  
  if(sum!=1){
    ans = ad(ans,ok);
  }
  return sum;
}

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);

  cin >> n;
  
  pwall=pw(2,n-1);
  rep(i,n-1){
    int a,b;
    cin >> a >> b;
    --a;--b;
    g[a].pb(b);
    g[b].pb(a);
  }
  dfs(0,-1);
  //  cout << "場合の数　" << ans << endl;
  cout << dv(ans,pw(2,n)) << endl;

  
  //  cout << endl << endl;
  //  cout << static_cast<double>(clock()) / CLOCKS_PER_SEC << "s"<<endl;
  return 0;
    

}
