//3~Aの内、0個以上の2,3,5,17,257,65537で割り切ることができる数
#include <iostream>
using namespace std;
int Z[6] = {2,3,5,17,257,65537};
int main(){
    int n;cin>>n;
    //0~A - 0~2
    int ans = 0;
    for(int bit = 1; (1<<6) > bit; bit++){
        int popcnt = __builtin_popcount(bit);
        int mul = 1;
        for(int i = 0; 6 > i; i++){
            if(bit >> i&1)mul *= Z[i];
        }
        if(popcnt%2)ans += n/mul;
        else ans -= n/mul;
    }
    cout << ans-1 << endl;
}