#pragma GCC optimization ("O3") #include using namespace std; using ll = long long; using vec = vector; using mat = vector; using pll = pair; using dvec = vector; using dmat = vector; #define INF (1LL<<61) //#define MOD 1000000007LL #define MOD 998244353LL #define EPS (1e-10) #define PR(x) cout << (x) << endl #define PS(x) cout << (x) << " " #define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i)) #define FORE(i,v) for(auto (i):v) #define ALL(x) (x).begin(), (x).end() #define SZ(x) ((ll)(x).size()) #define REV(x) reverse(ALL((x))) #define ASC(x) sort(ALL((x))) #define DESC(x) {ASC((x)); REV((x));} #define BIT(s,i) (((s)>>(i))&1) #define pb push_back #define fi first #define se second template inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;} template inline int chmax(T& a, T b) {if(a=MOD) x-=MOD; return *this;} mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;} mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;} mint operator+(const mint& a) const {mint b(*this); return b+=a;} mint operator-(const mint& a) const {mint b(*this); return b-=a;} mint operator*(const mint& a) const {mint b(*this); return b*=a;} mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;} mint inv() const {return pow(MOD-2);} mint& operator/=(const mint& a) {return *this*=a.inv();} mint operator/(const mint& a) const {mint b(*this); return b/=a;} }; istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;} ostream &operator<<(ostream& os, const mint& a) {return os<; using mmat = vector; vec sieve(ll n) { vec p(n+1); REP(i,0,n+1) p[i] = i; for(ll i=2; i*i<=n; ++i) { if(p[i] < i) continue; for(ll j=i*i; j<=n; j+=i) { if(p[j] == j) p[j] = i; } } return p; } int main() { ll N; cin >> N; vec R(N+1); for(ll i=1; i*i<=N; ++i) R[i*i] = 1; REP(i,1,N+1) R[i] += R[i-1]; ll M = R[N]; vec S = sieve(M), P; REP(i,2,M+1) { if(S[i] == i) P.pb(i); } vec F(N+1); REP(i,1,N+1) F[i] = i; FORE(p,P) { ll t = p*p; for(ll i=t; i<=N; i+=t) { while(F[i]%t == 0) F[i] /= t; } } ll ans = 0; REP(i,1,N+1) ans += R[N/F[i]]; PR(ans); return 0; } /* */