CF1183 F
I'm going to give you n numbers. I'm going to ask you to select at most 3 numbers. The numbers that don't divide each other make the sum the maximum.

The method is to enumerate a selected number x from large to small
And then he never divides all his numbers by the largest y
And then never divide the two numbers x and y by the largest number z.
Why do you have to choose the biggest one?
From this point of view
If the maximum number m is not a multiple of the next largest number m, then you can take the maximum and the second largest directly.
If the largest number m is a multiple of the next largest number m, then the smallest multiple 2
Secondary large + secondary large < secondary large + secondary large < = maximum

So choose the biggest

And then discretize the same thing, because the same things are bound to divide each other.

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
const int MAX_N = 200025;
int arr[MAX_N];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,ans = 0;
        scanf("%d",&n);
        for(int i = 1;i<=n;++i) scanf("%d",&arr[i]);
        sort(arr+1,arr+1+n);
        int sz = unique(arr+1,arr+1+n)-arr-1;
        for(int i = sz;i>=1;--i)
        {
            ans = max(ans,arr[i]);
            for(int j = i-1;j>=1;--j)
            {
                if(arr[i]%arr[j]!=0)
                {
                    ans = max(ans,arr[i]+arr[j]);
                    for(int k = j-1;k>=1;--k)
                    {
                        if(arr[i]%arr[k]!=0&&arr[j]%arr[k]!=0)
                        {
                            ans = max(ans,arr[i]+arr[j]+arr[k]);
                            break;
                        }
                    }
                    break;
                }
            }
        }
        printf("%d\n",ans);
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}