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)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

#define N 998244353

int a[200000];

unsigned long long int lcm(unsigned long long int a, unsigned long long int b)
{
	unsigned long long int x = a * b;
	unsigned long long int t, r;

	if (a < b)
	{
		t = a;
		a = b;
		b = t;
	}

	r = a % b;

	while (r!= 0)
	{
		a = b;
		b = r;
		r = a % b;
	}

	return x / b;
}

int main(void)
{
	int n;
	unsigned long long int t, ans = 0;

	scanf("%d", &n);

	for (int i = 0; i < n; i++)
	{
		scanf("%d", &t);
		a[i] = t;
	}

	for (int i = 0; i < n - 1; i++)
	{
		for (int j = i + 1; j < n; j++)
		{
			ans = (ans + lcm(a[i], a[j])) % N;
		}
	}
	
	printf("%d", ans);


	return 0;
}