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: Wrong Answer)

#include<cstdio>
#include<algorithm>
#include<map>
using namespace std;
#define N 200002
#define mod 998244353
#define M 1000000
#define ll long long
int a[N],n;
ll ans;
bool vis[M+5];
int gcd(int a,int b)
{
	if(!b)return a;
	return gcd(b,a%b);
}
int main()
{
	//freopen("number.in","r",stdin);
    //freopen("number.out","w",stdout);
    vis[4]=vis[6]=vis[8]=1;
    for(int i=3;i<=M;i+=2)
    	if(!vis[i])
    	{
    		//printf("%d\n",i);
    		for(ll j=1ll*i*i;j<=M;j+=i)vis[j]=1;
		}
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
		scanf("%d",&a[i]);
		if(!vis[a[i]])
		{
			for(int j=1;j<i;j++)
				if(a[j]%a[i]==0)ans=(ans+a[i])%mod;
				else ans=(ans+1)%mod;
		}
		else for(int j=1;j<i;j++)
			{
				if(!vis[a[j]])
				{
					if(a[i]%a[j]==0)ans=(ans+a[j])%mod;
					else ans=(ans+1)%mod;
				}
				else 
				{
				if(a[j]>a[i])ans=(ans+1ll*a[i]/gcd(a[j],a[i])*a[j])%mod;
				else ans=(ans+1ll*a[i]/gcd(a[i],a[j])*a[j])%mod;
				}
			}
   	}
   	printf("%lld",ans);
}