#pragma GCC optimization ("O3")

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pll = pair<ll,ll>;
using dvec = vector<double>;
using dmat = vector<dvec>;

#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<class T> inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;}
template<class T> inline int chmax(T& a, T b) {if(a<b) {a=b; return 1;} return 0;}
class mint {
public:
    ll x;
    mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
    mint operator-() const {return mint(-x);}
    mint& operator+=(const mint& a) {if((x+=a.x)>=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<<a.x;}
using mvec = vector<mint>;
using mmat = vector<mvec>;

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;  
}

/*



*/