#include using namespace std; using std::cout; using std::cin; using std::endl; using ll=long long; using ld=long double; const ll ILL=2167167167167167167; const int INF=2100000000; const int mod=998244353; #define rep(i,a,b) for (int i=(int)(a);i<(int)(b);i++) #define all(p) p.begin(),p.end() template using _pq = priority_queue, greater>; template ll LB(vector &v,T a){return lower_bound(v.begin(),v.end(),a)-v.begin();} template ll UB(vector &v,T a){return upper_bound(v.begin(),v.end(),a)-v.begin();} template bool chmin(T &a,T b){if(a>b){a=b;return 1;}else return 0;} template bool chmax(T &a,T b){if(a void So(vector &v) {sort(v.begin(),v.end());} template void Sore(vector &v) {sort(v.begin(),v.end(),[](T x,T y){return x>y;});} bool yneos(bool a,bool upp=0){if(a){cout<<(upp?"YES\n":"Yes\n");}else{cout<<(upp?"NO\n":"No\n");}return a;} template void vec_out(vector &p,int ty=0){ if(ty==2){cout<<'{';for(int i=0;i<(int)p.size();i++){if(i){cout<<",";}cout<<'"'< T vec_min(vector &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmin(ans,x);return ans;} template T vec_max(vector &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmax(ans,x);return ans;} template T vec_sum(vector &a){T ans=T(0);for(auto &x:a) ans+=x;return ans;} int pop_count(long long a){int res=0;while(a){res+=(a&1),a>>=1;}return res;} template bool inside(T l,T x,T r){return l<=x&&x Divisors(long long N){ vector p,q; long long i=1,K=0; while(i*i=0;i--){ p.push_back(q[i]); } return p; } void solve(); // CYAN / FREDERIC int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t = 1; // cin >> t; rep(i, 0, t) solve(); } void solve(){ ll N; cin >> N; auto D = Divisors(N); int L = D.size(); vector dp1(L), dp2(L); vector C(L); rep(i, 0, L) C[i] = N / D[i]; for (int i = L - 1; i >= 0; i--){ rep(j, i + 1, L) if (D[j] % D[i] == 0){ C[i] -= C[j]; } } map m; for (auto x : D) m[x] = m.size(); dp2[L - 1] = 1; for (int i = L - 1; i > 0; i--){ vector p(L); rep(j, 0, L) p[m[gcd(D[j], D[i])]] += C[j]; dp1[i] += dp2[i] * ((ld)(N) / (ld)(N - p[i])); rep(j, 0, i){ ld pro = (ld)(p[j]) / (ld)(N - p[i]); dp1[j] += dp1[i] * pro; dp2[j] += dp2[i] * pro; } } cout << fixed << setprecision(20) << dp1[0] << "\n"; }