Problem:
Score : 700 points

Problem Statement

We have an integer sequence of length N: A_0,A_1,\cdots,A_{N-1}.

Find the following sum (\mathrm{lcm}(a, b) denotes the least common multiple of a and b):

\sum_{i=0}^{N-2} \sum_{j=i+1}^{N-1} \mathrm{lcm}(A_i,A_j)

Since the answer may be enormous, compute it modulo 998244353.

Constraints

1 \leq N \leq 200000

1 \leq A_i \leq 1000000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_0\ A_1\ \cdots\ A_{N-1}

Output

Print the sum modulo 998244353.

Sample Input 1

3
2 4 6

Sample Output 1

22

\mathrm{lcm}(2,4)+\mathrm{lcm}(2,6)+\mathrm{lcm}(4,6)=4+6+12=22.

Sample Input 2

8
1 2 3 4 6 8 12 12

Sample Output 2

313

Sample Input 3

10
356822 296174 484500 710640 518322 888250 259161 609120 592348 713644

Sample Output 3

353891724

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
//dengyaotriangle!

const int maxn=200005;
const int maxm=1000005;
const int mdn=998244353;
const int inv2=(mdn+1)>>1;
int n;
int a[maxn];
int c[maxm];
long long f[maxm];
int mu[maxm];
bool ntp[maxm];
int prm[maxm],prc;
int cnt;
map<int,int> h;
int ex[maxn],px[maxn];
vector<int> fac[maxm];
void seive(){
    mu[1]=ntp[1]=1;
    for(int i=2;i<maxm;i++){
        if(!ntp[i]){prm[++prc]=i;mu[i]=-1;}
        for(int j=1;prm[j]*i<maxm;j++){
            ntp[i*prm[j]]=1;
            if(i%prm[j]){
                mu[i*prm[j]]=-mu[i];
            }else{
                mu[i*prm[j]]=0;
                break;
            }
        }
    }
}

inline long long calc(int n){
    long long tv=0;
    for(int i=0;i<fac[n].size();i++){
        tv=(tv+mu[fac[n][i]]*fac[n][i])%mdn;
    }
    return (tv+mdn)%mdn;
}
int main(){
    ios::sync_with_stdio(0);
    seive();
    cin>>n;
    long long ans=0,sm=0;
    for(int i=1;i<=n;i++){cin>>a[i];sm+=a[i];sm%=mdn;}
    for(int i=1;i<=n;i++){c[a[i]]++;}
    for(int i=1;i<maxm;i++){
        long long t=0;
        for(int j=1;j<=(maxm-1)/i;j++){
            t+=(long long)(j)*c[j*i];t%=mdn;
            fac[j*i].push_back(i);
        }
        f[i]=(t*t)%mdn;
    }
    for(int i=1;i<maxm;i++){
        long long sum=calc(i);//sum{d|n}mu[d]*d
        ans+=f[i]*i%mdn*sum%mdn;
        ans%=mdn;
    }
    cout<<((ans-sm+mdn)*inv2)%mdn;
    return 0;
}