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 ;

const long long mod = 1e9 + 7ll ;
const int MAX = 2e5 + 10 ;

long long sz[MAX] , P[MAX] ;
int n ;

long long ans = 0ll ;

vector< vector<int> >adj(MAX) ;

long long powlog(long long base , long long power)
{
	if(power == 0)
		return 1ll ;
	long long x = powlog(base , power / 2ll) ;
	x = (x * x) % mod ;
	if(power & 1)
		x = (x * base) % mod ;
	x %= mod ;
	return x ;
}

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

void dfs2(int node , int par)
{
	vector<long long>v ;
	long long sum = 0ll ;
	for(auto &child : adj[node])
	{
		if(child == par)
			continue ;
		dfs2(child , node) ;
		v.push_back(sz[child]) ;
		sum += sz[child] ;
	}
	long long x = (P[sz[node] - 1] - 1ll + mod) % mod ;
	long long y = (P[n - sz[node]] - 1ll + mod) % mod ;
	long long z = (x * y) % mod ;
	ans = (ans + z) % mod ;
	for(auto &i : v)
	{
		sum -= i ;
		x = (P[i] - 1ll + mod) % mod ;
		y = (P[sum] - 1ll + mod) % mod ;
		z = (x * y) % mod ;
		ans = (ans + z) % mod ;
	}
}

int main()
{
	ios_base::sync_with_stdio(0) ;
	cin.tie(0) ;
	P[0] = 1ll ;
	for(int i = 1 ; i < MAX ; ++i)
		P[i] = (P[i-1] * 2ll) % mod ;
	cin>>n ;
	for(int i = 0 ; i < n-1 ; ++i)
	{
		int x , y ;
		cin>>x>>y; 
		adj[x].push_back(y) ;
		adj[y].push_back(x) ;
	}
	dfs(1 , -1) ;
	dfs2(1 , -1) ;
	ans = (ans * powlog(P[n] , mod-2)) % mod ;
	ans %= mod ;
	cout<<ans<<"\n" ;
	return 0 ;
}		