There are n people standing in a line. Each of them has a unique id number.

Now the Ragnarok is coming. We should choose 3 people to defend the evil. As a group, the 3 people should be able to communicate. They are able to communicate if and only if their id numbers are pairwise coprime or pairwise not coprime. In other words, if their id numbers are a, b, c, then they can communicate if and only if [(a, b) = (b, c) = (a, c) = 1] or [(a, b) ≠ 1 and (a, c) ≠ 1 and (b, c) ≠ 1], where (x, y) denotes the greatest common divisor of x and y.

We want to know how many 3-people-groups can be chosen from the n people.

The first line contains an integer T (T ≤ 5), denoting the number of the test cases.

For each test case, the first line contains an integer n(3 ≤ n ≤ 10⁵), denoting the number of people. The next line contains n distinct integers a₁, a₂, . . . , a_n(1 ≤ a_i ≤ 10⁵) separated by a single space, where a_i stands for the id number of the i-th person.

For each test case, output the answer in a line.

1
5
1 3 9 10 2

4

/**
 * @author neko01
 */
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair<int,int> pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x7fffffff;
const LL inf=(((LL)1)<<61)+5;
const int N=100005;
const int M=500;
bool isprime[M];
int prime[M];
int tot=0,n;
int fac[50];
bool have[N];
int a[N];
int s[N];  //sum[i]表示a[i]中能整除i的个数
void getprime()
{
    for(int i=2;i<=M;i++)
    {
        if(isprime[i]) continue;
        prime[tot++]=i;
        for(int j=i*i;j<=M;j+=i)
            isprime[i]=true;
    }
}
LL gao()
{
    LL ans=0;
    for(int i=0;i<n;i++)
    {
        int m=a[i],num=0;
        LL sum=0;
        for(int j=0;j<tot&&prime[j]*prime[j]<=m;j++)
        {
            if(m%prime[j]==0)
            {
                fac[num++]=prime[j];
                while(m%prime[j]==0)
                    m/=prime[j];
            }
        }
        if(m!=1) fac[num++]=m;
        for(int j=1;j<(1<<num);j++)
        {
            int op=0,tmp=1;
            for(int k=0;k<num;k++)
            {
                if((1<<k)&j)
                {
                    tmp*=fac[k];
                    op++;
                }
            }
            if(op&1) sum+=s[tmp];
            else sum-=s[tmp];
        }
        if(sum==0) continue;
        ans+=(sum-1)*(n-sum);  //sum不互质的数包括了a[i]自身故减1
    }
    return ans/2;
}
int main()
{
    int t;
    scanf("%d",&t);
    getprime();
    while(t--)
    {
        scanf("%d",&n);
        clr(have);
        clr(s);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            have[a[i]]=true;
        }
        for(int i=2;i<N;i++)
            for(int j=i;j<N;j+=i)
                if(have[j]) s[i]++;
        LL ans=(LL)n*(n-1)*(n-2)/6;
        ans-=gao();
        printf("%I64d\n",ans);
    }
    return 0;
}