loj6229 这是一道简单的数学题

https://loj.ac/problem/6229

题解：https://blog.csdn.net/Vectorxj/article/details/79094659

套路推式子，杜教筛，证明复杂度。

感谢NicoDafaGood，不在这里写题解了，式子列出来太长啦，写在本本上。

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<map>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const ll mod=1e9+7;
const int maxn=2e6+7,W=2e6;
ll n,g[maxn],inv6;

char cc;ll ff;
template<typename T>void read(T& aa) {
	aa=0;cc=getchar();ff=1;
	while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

ll pf(ll x) {return x*x%mod;}

ll qp(ll x,ll k) {
	ll rs=1;
	while(k) {
		if(k&1) rs=rs*x%mod;
		k>>=1; x=x*x%mod;
	}
	return rs;
}

ll mu[maxn],prime[maxn],totp;
bool ok[maxn];
void get_p() {
	mu[1]=1;
	For(i,2,W) {
		if(!ok[i]) {
			prime[++totp]=i;
			mu[i]=-1;
		}
		For(j,1,totp) {
			if(prime[j]>W/i) break;
			ok[i*prime[j]]=1;
			if(i%prime[j]) mu[i*prime[j]]=-mu[i];
			else {
				mu[i*prime[j]]=0;
				break;
			}
		}
	}
	For(i,1,W) g[i]=((g[i-1]+mu[i]*pf(i))%mod+mod)%mod;
}

inline ll f2(ll x) {return x*(x+1)%mod*(2*x+1)%mod*inv6%mod;}

map<ll,ll> G,H;

ll get_g(ll n) {
	if(n<=W) return g[n];
	ll p=G[n]; if(p) return p;
	ll rs=0,lst=1,now;
	for(ll i=2,j;i<=n;i=j+1) {
		j=n/(n/i); now=f2(j);
		rs+=get_g(n/i)*(now-lst+mod)%mod;
		lst=now;
	}
	rs=(1-rs%mod+mod)%mod;
	return G[n]=rs;
}

inline ll get_h(ll n) {
	if(H[n]) return H[n];
	ll rs=0,lst=0,now;
	for(ll i=1,j;i<=n;i=j+1) {
		j=n/(n/i);
		now=get_g(j);
		rs+=(n/i)*(now-lst+mod)%mod;
		lst=now;
	}
	return H[n]=rs%mod;
}

ll get_f(ll n) {
	ll rs=0,x;
	for(ll i=1,j;i<=n;i=j+1) {
		j=n/(n/i); x=n/i;
		rs+=pf(((x+1)*x/2)%mod)*(get_h(j)-get_h(i-1)+mod)%mod;
	}
	return rs%mod;
}

int main() {
	read(n);
	get_p(); inv6=qp(6,mod-2);
	ll x=get_f(n);
	printf("%lld\n",(x+n)*qp(2,mod-2)%mod);
	return 0;
}