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
#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

const ll mod = 1e9 + 7;
ll modpow(ll A, ll B) {
	ll kotae = 1;
	while (B > 0) {
		if (B & 1) kotae = kotae * A % mod;
		A = A * A % mod;
		B >>= 1;
	}
	return kotae;
}

ll kotae = 0;
int N;
vector<int> E[200001];
int dfs(int v, int prev) {
	int kazu = 1;
	int honsuu = 0;

	for (int to : E[v]) {
		if (to != prev) {
			int k = dfs(to, v);
			kazu += k;
			honsuu++;
			kotae += modpow(2, k) - 1;
			kotae %= mod;
		}
	}
	if (honsuu == 0) {
		kotae += modpow(2, N - 1);
	}
	else if (prev == 0 && honsuu == 1) {
		kotae += 1;
	}
	else {
		kotae += 1;
		if (prev) {
			int k = N - kazu;
			kotae += modpow(2, k) - 1;
			kotae %= mod;
		}
	}
	kotae = (kotae + mod) % mod;

	return kazu;
}

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


	cin >> N;
	rep(i, N - 1) {
		int a, b;
		cin >> a >> b;
		E[a].pb(b);
		E[b].pb(a);
	}
	dfs(1, 0);
	ce(kotae);
	ll igai = modpow(2, N - 1) * N % mod;
	kotae = (igai - kotae + mod) % mod;


	ll waru = modpow(2, N);
	kotae = kotae * modpow(waru, mod - 2) % mod;

	co(kotae);

	Would you please return 0;
}