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;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl







using ll = long long;
const ll mod = 1e9 + 7;
const int maxn = 1e6 + 5;


ll pw(ll x, ll y) {
    ll res=1;
    while (y) {
	if (y%2) res=res*x%mod;
	x=x*x%mod;y=y/2;
    }
    return ((res%mod)+mod)%mod;
}

int n;
vector<int> g[maxn];

int par[maxn];
int siz[maxn];
vector<int> subtrees[maxn];

void dfs(int at, int p=0) {
    siz[at]=1;
    for (int to: g[at]) {
	if (to == p) continue;
	dfs(to, at);
	par[to] = at;
	siz[at] += siz[to];
    }
}


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

    cin>>n;
    for (int i=0; i<n-1; i++) {
	int u,v; cin>>u>>v;
	g[u].push_back(v);
	g[v].push_back(u);
    }

    dfs(1);

    
    for (int at=1; at<=n; at++) {
	ll tot = 0;
	for (int to: g[at]) {
	    if (to==par[at]) continue;
	    tot += siz[to];
	    subtrees[at].push_back(siz[to]);
	}
	ll up = n-1-tot;
	if (up>0) subtrees[at].push_back(up);
    }


    ll res = 0;
    for (int i=1; i<=n; i++) {
	ll all = pw(2,n-1);
	all --;
	for (ll x: subtrees[i]) {
	    all -= (pw(2,x)-1);
	    all %= mod;
	}
	res += all;
	res %= mod;
    }

    res %= mod;
    res += mod;
    res %= mod;

    ll ans = res*pw(pw(2,n),mod-2)%mod;
    cout<<ans<<endl;
    
    return 0;
}
