#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define Rep(b, e, i) for(int i = b; i <= e; i++) #define Repr(e, b, i) for(int i = e; i >= b; i--) #define rep(n, i) Rep(0, n-1, i) #define repr(n, i) Repr(n-1, 0, i) #define all(v) (v).begin(), (v).end() #define pb(v) push_back(v) #define uniq(v) (v).erase(unique(all(v)),(v).end()) #define bitcnt(x) __builtin_popcount(x) #define fst first #define snd second #define Pqaz(T) priority_queue,greater> #define Pqza(T) priority_queue #define put(x) cout << x; #define putsp(x) cout << x << ' '; #define putbd cout << "---------------------------------------------" << endl; #define putln(x) cout << x << endl; #define debug(x) cerr << #x << "=" << x << endl; #define ENJYU std::ios::sync_with_stdio(false);std::cin.tie(0); typedef long long ll; typedef pair llP; typedef pair intP; typedef complex comp; typedef vector vec; typedef vector vecll; typedef vector vecd; typedef vector mat; typedef vector matll; typedef vector matd; //vector の中身を出力 template ostream &operator<<(ostream &o,const vector&v) {o<<"{";for(int i=0;i<(int)v.size();i++)o<<(i>0?", ":"")< Vec; typedef vector Mat; //A * B = C Mat mul(const Mat &A, const Mat &B) { //大前提 assert((int)A[0].size() == (int)B.size()); Mat C((int)A.size(), Vec((int)B[0].size(), plus_base)); //メモリへのアクセス的にこうしたほうが早いのです rep((int)A.size(), i) rep((int)B.size(), k) rep((int)B[0].size(), j) { C[i][j] += A[i][k] * B[k][j] % MOD; C[i][j] %= MOD; } return C; } // v' = A * v //オーバーロードしているんです!!!!!!!!!!!! Vec mul(const Mat &A, const Vec &v) { //大前提 assert((int)A[0].size() == (int)v.size()); Vec nv((int)A.size(), plus_base); rep((int)A.size(), i) rep((int)v.size(), j) { nv[i] += A[i][j] * v[j] % MOD; nv[i] %= MOD; } return nv; } // A ^ N Mat mpow(Mat A, int N) { //大前提 assert((int)A[0].size() == (int)A.size() && N > 0); Mat Res((int)A.size(), Vec((int)A.size(), plus_base)); // A ^ 0 = E rep((int)A.size(), i) { Res[i][i] = mul_base; } while(N) { if (N & 1) { Res = mul(Res, A); } A = mul(A, A); N >>= 1; } return Res; } void solve(void) { ll N; cin >> N; Mat m(2, Vec(2, 1LL)); m[1][1] = 0LL; m = mpow(m, N); ll ans = m[0][0] * m[0][1] % MOD; cout << ans << endl; } int main(void){ solve(); //cout << "yui(*-v・)yui" << endl; return 0; }