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: Wrong Answer)

#include <stdio.h>
#define N 200000
int A[N];

unsigned long long int gcd(int x, int y){
  while(x != 0){
    int z = x;
    x = y%x;
    y = z;
  }
  return y;
} 

unsigned long long int lcm(int x, int y){
  int z = gcd(x, y);
  int a = x*y/z;
  return a;
}

unsigned long long int main(){
  int n,i,j,z;
  unsigned long long int sum, r;
  sum = 0;
  for(i=0; i<=n-2; i++){
    for(j=i+1; j<=n-1; j++){
      z = lcm(A[i], A[j]);
      sum = sum + z;
    }
  }
  r = sum % 998244353;
  printf("%d\n", r);
  return 0;
}