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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

int main(){
  
  long long n,b,d,e;
  long long an=0;
  scanf("%d",&n);
  long long a[n];
  for(long long i=0;i<n;i++){
    scanf("%lld",&a[i]);
  }
  for(long long i=0;i<n-1;i++){
    for(long long j=i+1;j<n;j++){
      b=1;
      long long c=a[i],d=a[j];
      if (c>d){
        e=c/2;
      }else{
        e=d/2;
      }
      for(long long k=2;k<e+1;k++){
        while(c%k==0&&d%k==0){
          c/=k;
          d/=k;
          b*=k;
        }
      }
      an+=c*d*b;
    }
  }
  an=an%998244353;
  if (an<0)an+=998244353;
  printf("%lld\n",an);
  return 0;
}