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 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<stdio.h>
#include<vector>
#include<queue>
#include<math.h>
#include<deque>
using namespace std;
#define int long long
#define rep(s,i,n) for(int i=s;i<n;i++)
#define c(n) cout<<n<<endl;
#define ic(n) int n;cin>>n;
#define sc(s) string s;cin>>s;
#define mod 998244353
#define inf 1000000000000000007
#define f first
#define s second
#define mini(c,a,b) *min_element(c+a,c+b)
#define maxi(c,a,b) *max_element(c+a,c+b)
#define pi 3.141592653589793238462643383279
#define e_ 2.718281828459045235360287471352
#define P pair<int,int>
#define upp(a,n,x) upper_bound(a,a+n,x)-a;
#define low(a,n,x) lower_bound(a,a+n,x)-a;
#define UF UnionFind 
#define pb push_back
//printf("%.12Lf\n",);
int keta(int x) {
	rep(0, i, 30) {
		if (x < 10) {
			return i + 1;
		}
		x = x / 10;
	}
}
int gcd(int x, int y) {
	if (x == 0 || y == 0)return x + y;
	int aa = x, bb = y;
	rep(0, i, 1000) {
		aa = aa % bb;
		if (aa == 0) {
			return bb;
		}
		bb = bb % aa;
		if (bb == 0) {
			return aa;
		}
	}
}
int lcm(int x, int y) {
	int aa = x, bb = y;
	rep(0, i, 1000) {
		aa = aa % bb;
		if (aa == 0) {
			return x / bb * y;
		}
		bb = bb % aa;
		if (bb == 0) {
			return x / aa * y;
		}
	}
}
bool prime(int x) {
	if (x == 1)return false;
	rep(2, i, sqrt(x) + 1) {
		if (x % i == 0 && x != i) {
			return false;
		}
	}
	return true;
}
int max(int a, int b) {
	if (a >= b)return a;
	else return b;
}
string maxst(string s, string t) {
	int n = s.size();
	int m = t.size();
	if (n > m)return s;
	else if (n < m)return t;
	else {
		rep(0, i, n) {
			if (s[i] > t[i])return s;
			if (s[i] < t[i])return t;
		}
		return s;
	}
}
string minst(string s, string t) {
	int n = s.size();
	int m = t.size();
	if (n < m)return s;
	else if (n > m)return t;
	else {
		rep(0, i, n) {
			if (s[i] < t[i])return s;
			if (s[i] > t[i])return t;
		}
		return s;
	}
}
string string_reverse(string s){
	int n=s.size();
	string t;
	rep(0,i,n)t+=s[n-i-1];
	return t;
}	
int min(int a, int b) {
	if (a >= b)return b;
	else return a;
}
int n2[70];
int nis[70];
int nia[70];
int mody[70];
int nn;
int com(int n, int y) {
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	int bunsi = 1, bunbo = 1;
	rep(0, i, y)bunsi = (bunsi * (n - i)) % mod;
	rep(0, i, y)bunbo = (bunbo * (i + 1)) % mod;
	mody[0] = bunbo;
	rep(1, i, 41) {
		bunbo = (bunbo * bunbo) % mod;
		mody[i] = bunbo;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = mod - 2;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn > n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	nis[0]++;
	rep(0, i, 41) {
		if (nis[i] == 1) {
			bunsi = (bunsi * mody[i]) % mod;
		}
	}
	return bunsi;
}
int gyakugen(int n, int y) {
	int ni = 1;
	for (int i = 0;i < 31;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	mody[0] = y;
	rep(1, i, 31) {
		y = (y * y) % mod;
		mody[i] = y;
	}
	rep(0, i, 31)nis[i] = 0;
	nn = mod - 2;
	for (int i = 30;i >= 0;i -= 1) {
		if (nn > n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	nis[0]++;
	rep(0, i, 31) {
		if (nis[i] == 1) {
			n = (n * mody[i]) % mod;
		}
	}
	return n;
}
int yakuwa(int n) {
	int sum = 0;
	rep(1, i, sqrt(n + 1)) {
		if (n % i == 0)sum += i + n / i;
		if (i * i == n)sum -= i;
	}
	return sum;
}
int poow(int y, int n) {
	if (n == 0)return 1;
	n -= 1;
	int ni = 1;
	for (int i = 0;i < 62;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	int yy = y;
	mody[0] = yy;
	rep(1, i, 62) {
		yy = (yy * yy) % mod;
		mody[i] = yy;
	}
	rep(0, i, 62)nis[i] = 0;
	nn = n;
	for (int i = 62;i >= 0;i -= 1) {
		if (nn >= n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	rep(0, i, 62) {
		if (nis[i] == 1) {
			y = (y * mody[i]) % mod;
		}
	}
	return y;
}
int minpow(int x, int y) {
	int sum = 1;
	rep(0, i, y)sum *= x;
	return sum;
}
int ketawa(int x, int sinsuu) {
	int sum = 0;
	rep(0, i, 80){
		if(minpow(sinsuu,i)>x){
          return sum;
        } 
		sum += (x % minpow(sinsuu, i + 1)) / (minpow(sinsuu, i));
	}	
	return sum;
}
double distance(double a,double b,double c,double d){
	return sqrt((b-a)*(b-a)+(c-d)*(c-d));
}	
int sankaku(int a) {
	return a * (a + 1) / 2;
}
int sames(int a[1111111], int n) {
	int ans = 0;
	rep(0, i, n) {
		if (a[i] == a[i + 1]) {
			int j = i;
			while (a[j + 1] == a[i] && j <= n - 2)j++;
			ans += sankaku(j - i);
			i = j;
		}
	}
	return ans;
}
using Graph = vector<vector<int>>;
int oya[114514];
int depth[114514];
void dfs(const Graph& G, int v, int p, int d) {
	depth[v] = d;
	oya[v] = p;
	for (auto nv : G[v]) {
		if (nv == p) continue; // nv が親 p だったらダメ
		dfs(G, nv, v, d + 1); // d を 1 増やして子ノードへ
	}
}
/*int H=10,W=10;
char field[10][10];
char memo[10][10];
void dfs(int h, int w) {
	memo[h][w] = 'x';

	// 八方向を探索
	for (int dh = -1; dh <= 1; ++dh) {
		for (int dw = -1; dw <= 1; ++dw) {
			if(abs(0-dh)+abs(0-dw)==2)continue;
			int nh = h + dh, nw = w + dw;

			// 場外アウトしたり、0 だったりはスルー
			if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
			if (memo[nh][nw] == 'x') continue;

			// 再帰的に探索
			dfs(nh, nw);
		}
	}
}*/
int XOR(int a, int b) {
	if (a == 0 || b == 0) {
		return a + b;
	}
	int ni = 1;
	rep(0, i, 41) {
		n2[i] = ni;
		ni *= 2;
	}
	rep(0, i, 41)nis[i] = 0;
	for (int i = 40;i >= 0;i -= 1) {
		if (a >= n2[i]) {
			nis[i]++;
			a -= n2[i];
		}
		if (b >= n2[i]) {
			nis[i]++;
			b -= n2[i];
		}
	}
	int sum = 0;
	rep(0, i, 41)sum += (nis[i] % 2 * n2[i]);
	return sum;
}
//int ma[1024577][21];
//for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;
struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
 
	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++) par[i] = i;
	}
 
	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}
 
	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}
 
	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};
int a[214514],cnt[1000011];
vector<int> l[1000011];
vector<int> yaku[1000011];
int ans[1000011];
signed main(){
	ic(n) 
	rep(0,i,n){
		cin>>a[i];
		cnt[a[i]]++;
	}	
	rep(1,i,1000001){
		int s=i;
		rep(1,j,1000001/i+1){
			yaku[s].pb(i);
			if(cnt[s]>0){
				rep(0,m,cnt[s])l[i].pb(s);
			}
			s+=i;
		}
	}
}