#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; using ll = long long; using P = pair; constexpr int INF = 1001001001; constexpr int mod = 1000000007; // constexpr int mod = 998244353; template inline bool chmax(T& x, T y){ if(x < y){ x = y; return true; } return false; } template inline bool chmin(T& x, T y){ if(x > y){ x = y; return true; } return false; } struct mint { int x; mint() : x(0) {} mint(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {} mint& operator+=(const mint& p){ if((x += p.x) >= mod) x -= mod; return *this; } mint& operator-=(const mint& p){ if((x -= p.x) < 0) x += mod; return *this; } mint& operator*=(const mint& p){ x = (int)(1LL * x * p.x % mod); return *this; } mint& operator/=(const mint& p){ *this *= p.inverse(); return *this; } mint operator-() const { return mint(-x); } mint operator+(const mint& p) const { return mint(*this) += p; } mint operator-(const mint& p) const { return mint(*this) -= p; } mint operator*(const mint& p) const { return mint(*this) *= p; } mint operator/(const mint& p) const { return mint(*this) /= p; } bool operator==(const mint& p) const { return x == p.x; } bool operator!=(const mint& p) const { return x != p.x; } mint pow(int64_t n) const { mint res = 1, mul = x; while(n > 0){ if(n & 1) res *= mul; mul *= mul; n >>= 1; } return res; } mint inverse() const { return pow(mod - 2); } friend ostream& operator<<(ostream& os, const mint& p){ return os << p.x; } friend istream& operator>>(istream& is, mint& p){ int64_t val; is >> val; p = mint(val); return is; } }; mint dp[1000005][4]; int main(){ ios::sync_with_stdio(false); cin.tie(nullptr); int N; cin >> N; dp[0][0] = 1; for(int i = 0; i < N; ++i){ for(int j = 0; j <= 3; ++j){ if(dp[i][j] == 0) continue; for(int k = 1; k <= 3; ++k){ if(k == j) continue; dp[i + k][k] += dp[i][j]; } } } cout << dp[N][1] + dp[N][2] + dp[N][3] << endl; return 0; }