#include <cstdio>
#include <vector>

using namespace std;

#define DxBG
#define FL

#define fin  "pairs.in"
#define fout "pairs.out"

#define pb push_back
#define sz(c) (int)((c).size())

const int Vmax = 1001100;
const int Nmax = 100100;

int N,MAX,v[Vmax];
int nr[Nmax];
vector <int> divnr[Nmax];

long long cnt1;	//cate is bune
long long cnt2;	//cate au cmmdc>1

//v[i]  - cate numere se divid cu i
//nr - numerele date
//div[i] - divizorii numarului nr[i]

void readdata()
{
	int i,a;

	freopen(fin,"r",stdin);

	scanf("%d",&N);

	for (i=1;i<=N;++i)
	{
		scanf("%d",&a);
		v[a]=1;
		if ( a > MAX )
			MAX = a;
		nr[i]=a;
	}
}

void solve()
{
	int i,j,k,nrb,tmp,bun;
	long long cnt;

//aflu fiecare numar "i" cu cate numere din "nr" se divide
	for (i=2;i<=MAX;++i)
	{
		for (j=i+i;j<Vmax;j+=i)
			if (v[j])
				++v[i];
	}
//gata

//aflu divizorii fiecarui numar
	for (i=1;i<=N;++i)
	{
		tmp=nr[i];
		for (j=2;j*j<=tmp;++j)
		{
			bun=0;
			while (tmp%j==0)
			{
				bun=1;
				tmp/=j;
			}
			if (bun)
				divnr[i].pb(j);
		}

		if (tmp>1)
			divnr[i].pb(tmp);
	}
//gata


#ifdef DBG
	for (i=1;i<=MAX;++i)
		printf("%d ",v[i]);
	printf("\n");
	for (i=1;i<=N;++i,printf("\n"))
	for (j=0;j<sz(divnr[i]);++j)
		printf("%d ",divnr[i][j]);
	printf("\n");
#endif


	cnt1 = N-1;
	cnt1 = cnt1 * N;
	cnt1 = cnt1 / 2;

	cnt2 = 0;

//incep rezolvarea propriu-zisa

	for (i=1;i<=N;++i)
	{
		cnt = 0;	//numar perechi care au cmmdc( nr[i] , a ) > 1

	//excludere si includere
		for (j=1;j<(1<<sz(divnr[i]));++j)
		{
			nrb = 0;
			tmp = 1;
			for (k=0;(1<<k)<=j;++k)
				if (j&(1<<k))
				{
					++nrb;
					tmp*=divnr[i][k];
				}
			if (nrb%2!=0)
				cnt+=(v[tmp]-1);
			else
				cnt-=(v[tmp]-1);
		}
	//gata
	
		cnt2+=cnt;

#ifdef DBG
		printf("%d %d\n",i,cnt);
#endif
	}

	cnt2/=2;

	cnt1-=cnt2;

#ifdef FL
	freopen(fout,"w",stdout);
#endif

	printf("%lld\n",cnt1);
}

int main()
{
	readdata();
	solve();
	return 0;
}