#include <bits/stdc++.h>
#define _GLIBCXX_DEBUG
using namespace std;
using ll = long long;
using V = vector<int>;
using VV = vector<V>;
#define rep2(i, m, n) for (int i = (m); i < (n); ++i)
#define drep2(i, m, n) for (int i = (m)-1; i >= (n); --i)
#define rep(i, n) rep2(i, 0, n)
#define drep(i, n) drep2(i, n, 0)
struct fast_ios { fast_ios(){ cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
const int MOD  = int(1e9)+7;


struct mint {
  ll x;
  mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) { if ((x += a.x) >= MOD) x -= MOD; return *this; }
  mint& operator-=(const mint a) { if ((x -= a.x) < 0) x += MOD; return *this; }
  mint& operator*=(const mint a) { (x *= a.x) %= MOD; return *this; }
  mint operator+(const mint a) const { return mint(*this) += a; }
  mint operator-(const mint a) const { return mint(*this) -= a; }
  mint operator*(const mint a) const { return mint(*this) *= a; }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
  mint& operator/=(const mint r) {
    ll a = r.x, b = MOD, u = 1, v = 0;
    while (b) {
      ll t = a / b;
      a -= t * b; swap(a, b);
      u -= t * v; swap(u, v);
    }
    x = x * u % MOD;
    if (x < 0) x += MOD;
    return *this;
  }
  mint operator/(const mint a) const { return mint(*this) /= a; }
};
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
using Vm = vector<mint>;


VV G;

Vm dfs(int k, int u, int p=-1) {
  Vm res = {1, 1, 0, 0};
  for (int v : G[u]) if (v != p) {
    auto q = dfs(k, v, u);
    q[0] *= 2;
    q[k] *= 2;
    drep(i, 4) {
      res[i] *= q[0];
      rep2(j, 1, i+1) res[i] += res[i-j] * q[j];
    }
  }
  return res;
}


int main() {
  int n; cin >> n;

  G.assign(n, V());
  rep(i, n-1) {
    int u, v; cin >> u >> v;
    --u, --v;
    G[u].push_back(v);
    G[v].push_back(u);
  }

  auto dp2 = dfs(2, 0);
  auto dp3 = dfs(3, 0);

  mint ans = mint(2).pow(n-1) * n;
  ans += dp2[2] * 6;
  ans += dp3[3] * 6;
  cout << ans << '\n';
  return 0;
}