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;
typedef long double ld;
const ll mod = 1e9 + 7;
const int N = 2e5 + 7;

vector<int> conn[N];
int subtree_size[N];
int parent[N];

int dfs(int n) {
	int sz = 1;
	for (auto i : conn[n]) {
		if (i == parent[n])
			continue;
		parent[i] = n;
		sz += dfs(i);
	}
	return subtree_size[n] = sz;
}

ll add(ll a, ll b) {
	return ((a%mod) + (b%mod)) % mod;
}

ll sub(ll a, ll b) {
	return ((a%mod) - (b%mod) + mod) % mod;
}

ll mul(ll a, ll b) {
	return ((a%mod) * (b%mod)) % mod;
}


ll fp(ll b, ll e) {
	if (!e)
		return 1;
	ll ans = fp(b, e / 2);
	ans = mul(ans, ans);
	if (e & 1) {
		ans = mul(b, ans);
	}
	return ans;
}

ll divide(ll a, ll b) {
	return ((a%mod) * fp(b, mod - 2)) % mod;
}



int main() {
	int n;
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		int u, v;
		cin >> u >> v;
		conn[u].push_back(v);
		conn[v].push_back(u);
	}
	dfs(1);
	ll ans = 0;
	for (int i = 1; i <= n; i++) {
		ll totalSz = 0;
		ll totalProb = 0;
		if (conn[i].size() < 2)
			continue;
		for (auto j : conn[i]) {
			if (j == parent[i])
				continue;
			ll prob = divide(sub(fp(2, subtree_size[j]), 1), fp(2, n));
			//cout << j << " " << subtree_size[j] << endl;
			totalProb = add(totalProb, prob);
			totalSz = add(totalSz, subtree_size[j]);
		}
		if (i > 1) {
			ll prob = divide(sub(fp(2, n - subtree_size[i]), 1), fp(2, n));
			totalProb = add(totalProb, prob);
		}
		totalProb = add(totalProb, divide(1, fp(2, n)));
		ans = add(ans, sub(divide(1, 2), totalProb));
	}
	cout << ans;
}