#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pii pair<int, int>
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep3(i, l, n) for (int i = l; i < (n); ++i)
#define sz(v) (int)v.size()
const int inf = 1e9 + 7;
const ll INF = 1e18;
#define abs(x) (x >= 0 ? x : -(x))
#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())
#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }
template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<typename T> T pow(T a, int b) { return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1; }
const int mod = 1000000007;
ll modpow(ll a, int b) { return b ? modpow(a * a % mod, b / 2) * (b % 2 ? a : 1) % mod : 1; }
template<class T> ostream& operator<<(ostream& os, const vector<T>& vec) { for (auto &vi: vec) os << vi << " "; return os; }
template<class T, class U> ostream& operator<<(ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }
template<class T, class T2> inline void add(T &a, T2 b) { a += b; if (a >= mod) a -= mod; }
void solve();
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int T;
// cin >> T;
T = 1;
while (T--) {
solve();
}
}
int n;
vector<vector<int> > g;
ll ans;
int dfs(int now, int prev) {
int sum = 1;
ll tmpans = 1; // すべて白
for (int nxt : g[now]) {
if (nxt == prev) continue;
int tmp = dfs(nxt, now);
sum += tmp;
add(tmpans, (modpow(2, tmp) - 1 + mod) % mod); // debuged % > +, -
}
add(tmpans, (modpow(2, n - sum) - 1 + mod) % mod);
tmpans = (modpow(2, n - 1) - tmpans + mod) % mod; // 余事象だから
if (sz(g[now]) != 1) add(ans, tmpans); // 葉だったら部分木 1 つ
return sum;
}
void solve() {
n = 200000;
// cin >> n;
g.resize(n);
rep(i, n - 1) {
int a, b;
a = i; b = i + 1;
// cin >> a >> b;
// a--; b--;
g[a].eb(b);
g[b].eb(a);
}
/*
各頂点が何回白として含まれるか
2 つ以上の部分木で 1 つ以上黒の頂点がある
余事象 : すべて白 or 1 つの部分木だけ黒を含む
すべて白 : 1 通り
1 つの... : 2^{部分木のサイズ} - 1
- 1 : 全部白は避ける, 他の部分木の頂点はすべて白
*/
dfs(0, -1);
(ans *= modpow(modpow(2, n), mod - 2) % mod) %= mod; // 期待値
cout << ans << endl;
}
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