#include <bits/stdc++.h>
using namespace std;
const int N = 1e7 + 10;
typedef long long LL;
int prime[N], pr, phi[N];
//prime存质数，pr代表现在筛出来了几个质数，phi存欧拉函数 
LL sp[N]; //欧拉函数前缀和 
bool v[N];
void init() {
    pr = 0; phi[1] = 1; //欧拉函数初始化，1就是 1 
    memset(v, 0, sizeof(v));
    for (int i = 2; i <= N; i++) {
        if (v[i] == 0) prime[++pr] = i, phi[i] = i - 1;
        for (int j = 1; (j <= pr) && (i * prime[j] <= N - 10); j++) {
            v[i * prime[j]] = 1;
            if (i % prime[j] == 0) {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            } else {
                phi[i * prime[j]] = phi[i] * (prime[j] - 1);                          
            }
        }
    }
    sp[0] = 0;
    for (int i = 1; i <= N - 10; i++) sp[i] = sp[i - 1] + phi[i];
} 
map<LL, LL> hs;
LL solve(LL x) {
    if (x <= N - 10) return sp[x];
    if (hs[x]) return hs[x];
    LL res = 0;
    for (LL i = 2, j; i <= x; i = j + 1) {
        j = x / (x / i);
        //分块加速 x/i都一样的 i分成一块，一块块的计算
        //这里 j是 x/j=x/i 中的最大的 j 
        res += (j - i + 1) * solve(x / i);
    }
    return hs[x] = x * (x + 1) / 2 - res;
}
int main() {
    init();
    LL n; scanf("%lld", &n);
    printf("%lld\n", solve(n));
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N=1e7+10;
typedef long long LL;
int prime[N], pr, phi[N];
//prime存质数，pr代表现在筛出来了几个质数，phi存欧拉函数 
LL sp[N]; //欧拉函数前缀和 
bool v[N];
void init(){
	pr=0; phi[1]=1; //欧拉函数初始化，1就是 1 
	memset(v, 0, sizeof(v));
	for(int i=2; i<=N; i++)
	{
		if(v[i]==0) prime[++pr]=i, phi[i]=i-1;
		for(int j=1; (j<=pr)&&(i*prime[j]<=N-10); j++)
		{
			v[i*prime[j]]=1;
			if(i%prime[j]==0){
				phi[i*prime[j]]=phi[i]*prime[j];
				break;
			}
			else{
				phi[i*prime[j]]=phi[i]*(prime[j]-1);                          
			}
		}
	}
	sp[0]=0;
	for(int i=1; i<=N-10; i++) sp[i]=sp[i-1]+phi[i];
} 
map<LL, LL> hs;
LL solve(LL x){
	if(x<=N-10) return sp[x];
	if(hs[x]) return hs[x];
	LL res=0;
	for(LL i=2, j; i<=x; i=j+1){
		j=x/(x/i);
		//分块加速 x/i都一样的 i分成一块，一块块的计算
		//这里 j是 x/j=x/i 中的最大的 j 
		res+=(j-i+1)*solve(x/i);
	}
	return hs[x]=x*(x+1)/2-res;
}
int main(){
	init();
	LL n; scanf("%lld",&n);
	printf("%lld\n", solve(n));
	return 0;
}