#include <iostream>
#include <vector>
using namespace std;
#pragma warning (disable: 4996)

const long long mod = 1000000007;
const long long inv2 = 500000004;

// 入出力
long long N, A[1 << 19], B[1 << 19];
long long Answer1, Answer2, Answer3A, Answer3B, Answer3C;
vector<int> X[1 << 19];
vector<int> Y[1 << 19];

// DP
int par[1 << 19];
long long dp1[1 << 19];
long long dp2[1 << 19];
long long dpr[1 << 19][4];
long long Eval[1 << 19];

void dfs1(int pos) {
	dp1[pos] += 1LL;
	for (int i : X[pos]) {
		if (dp1[i] != 0LL) continue;
		dfs1(i);
		par[i] = pos;
		Y[pos].push_back(i);
		dp1[pos] += dp1[i] * inv2 % mod;
		dp1[pos] %= mod;
	}
}

void dfs2(int pos) {
	for (int i : Y[pos]) {
		dp2[i] += dp2[pos] * inv2 % mod;
		dp2[i] += dp1[pos] * inv2 % mod;
		dp2[i] -= dp1[i] * (inv2 * inv2 % mod) % mod;
		dp2[i] = (dp2[i] + mod * mod) % mod;
		dfs2(i);
	}
}

void getEval() {
	for (int i = 1; i <= N; i++) {
		long long r = 0, s1 = 0, s2 = 0;
		for (int j : Y[i]) s1 += dp1[j] * inv2 % mod;
		for (int j : Y[i]) s2 += (dp1[j] * inv2 % mod) * (dp1[j] * inv2 % mod) % mod;
		s1 %= mod; s2 %= mod;
		r = s1 * s1 - s2;
		r = (r + mod * mod) % mod;
		Eval[i] = r * inv2 % mod;
	}
}

int main() {
	// Step #1. 入力
	scanf("%lld", &N);
	for (int i = 1; i <= N - 1; i++) scanf("%lld%lld", &A[i], &B[i]);
	for (int i = 1; i <= N - 1; i++) {
		X[A[i]].push_back(B[i]);
		X[B[i]].push_back(A[i]);
	}
	
	// Step #2. 全方位木DP
	dfs1(1);
	dfs2(1);

	// Step #3. [1], [2] の場合を求める
	Answer1 = N;
	getEval();
	for (int i = 1; i <= N; i++) Answer2 += (dp1[i] + mod - 1LL) % mod;
	for (int i = 1; i <= N; i++) Answer2 += Eval[i];
	Answer2 %= mod;

	// Step #4. [3] の場合 - Pattern A
	for (int i = 1; i <= N; i++) {
		long long r1 = (dp1[i] - 1LL);
		long long r2 = dp2[i];
		Answer3A += r1 * r2 % mod;
		Answer3A %= mod;
	}
	for (int i = 1; i <= N; i++) {
		Answer3A += Eval[i];
		Answer3A %= mod;
	}

	// Step #5. [3] の場合 - Pattern B
	for (int i = 1; i <= N; i++) {
		long long r1 = dp2[i];
		long long r2 = Eval[i];
		Answer3B += r1 * r2 % mod;
		Answer3B %= mod;
	}

	// Step #6. [3] の場合 - Pattern C
	for (int i = 1; i <= N; i++) {
		for (int j = 0; j <= Y[i].size(); j++) { dpr[j][0] = 0; dpr[j][1] = 0; dpr[j][2] = 0; dpr[j][3] = 0; }
		dpr[0][0] = 1;
		for (int j = 0; j < Y[i].size(); j++) {
			for (int k = 0; k <= 3; k++) dpr[j + 1][k] = dpr[j][k];
			for (int k = 0; k <= 2; k++) {
				dpr[j + 1][k + 1] += dpr[j][k] * (dp1[Y[i][j]] * inv2 % mod) % mod;
				dpr[j + 1][k + 1] %= mod;
			}
		}
		Answer3C += dpr[Y[i].size()][3];
		Answer3C %= mod;
	}

	// Step #7. 出力
	long long Ret1 = Answer1;
	long long Ret2 = Answer2;
	long long Ret3 = (Answer3A + Answer3B + Answer3C) % mod;
	for (int i = 1; i <= N - 1; i++) { Ret1 *= 2LL; Ret1 %= mod; }
	for (int i = 1; i <= N - 1; i++) { Ret2 *= 2LL; Ret2 %= mod; }
	for (int i = 1; i <= N - 1; i++) { Ret3 *= 2LL; Ret3 %= mod; }
	cout << (1LL * Ret1 + 6LL * Ret2 + 6LL * Ret3) % mod << endl;
	return 0;
}