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 pb push_back
#define se second
const int N = 2e5+5;
const int mod = 1e9+7;
ll po(int a, int b){
    if(b==0)return 1;
    ll x = po(a,b/2);
    x = (x*x)%mod;
    if(b&1)x = (x*a)%mod;
    return x;
}
ll z[N],n;
bool vis[N];
int par[N];
vector<int>v[N];
ll ans = 0;
void dfs(int x){
    vis[x] = true;
    z[x] = 1;
    for(int y:v[x]){
        if(vis[y])continue;
        par[y] = x;
        dfs(y);
        z[x]+=z[y];
    }
    if(z[x]==1)return;
    ll trt1 = po(2,z[x]-1)-1;
    ll trt2 = po(2,n-z[x])-1;
    trt1=(trt1*trt2)%mod;
    trt2 = po(2,n);
    trt2 = po(trt2,mod-2);
    trt1 = (trt1*trt2)%mod;
    ans = (ans+trt1)%mod;
    trt1 = 1;
    for(int y:v[x]){
        if(par[y]!=x)continue;
        trt1+=po(2,z[y])-1;
        trt1%=mod;
    }
    trt1 = ((po(2,z[x]-1)-trt1)%mod+mod)%mod;
    trt2 = po(2,n);
    trt2 = po(trt2,mod-2);
    trt1 = (trt1*trt2)%mod;
    ans = (ans+trt1)%mod;
}
void solve(){
    cin>>n;
    for(int i=1;i<n;i++){
        int x,y;
        cin>>x>>y;
        v[x].pb(y);
        v[y].pb(x);
    }
    dfs(1);
    cout<<ans<<endl;
}
int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t=1;
   // cin>>t;
    while(t--){
        solve();
    }
    return 0;
}
