#include #include #include #include #include #include #include #include #include #include #include #include using namespace std; using ll = long long; const ll modc = 1e9+7; class mint { ll x; public: mint(ll x=0) : x((x%modc+modc)%modc) {} mint operator-() const { return mint(-x); } mint& operator+=(const mint& a) { if ((x += a.x) >= modc) x -= modc; return *this; } mint& operator-=(const mint& a) { if ((x += modc-a.x) >= modc) x -= modc; return *this; } mint& operator*=(const mint& a) { (x *= a.x) %= modc; return *this; } mint operator+(const mint& a) const { mint res(*this); return res+=a; } mint operator-(const mint& a) const { mint res(*this); return res-=a; } mint operator*(const mint& a) const { mint res(*this); return res*=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 inv() const { return pow(modc-2); } mint& operator/=(const mint& a) { return (*this) *= a.inv(); } mint operator/(const mint& a) const { mint res(*this); return res/=a; } bool operator == (const mint& a) const{ return x == a.x; } friend ostream& operator<<(ostream& os, const mint& m){ os << m.x; return os; } friend istream& operator>>(istream& ip, mint &m) { ll t; ip >> t; m = mint(t); return ip; } }; int main(){ ll N; cin >> N; vector A(N), B(N+1); B[0] = 1; for (int i=0; i> A[i]; B[i+1] = (B[i] * A[i]); } mint ans = B[N]; for (int i=1; i<=N-1; i++) ans += B[i] * mint(3).pow(N-1-i) * 2; cout << ans << endl; }