Minimum Sum of Array（map)

You are given an array a consisting of n integers a₁, ..., a_n. In one operation, you can choose 2 elements a_i and a_j in which a_i is divisible by a_j and transform a_i to a_j.

A number x is said to be divisible by a number y if x can be divided by y and the result is an exact whole number. For example, 15 is divisible by 3, because 15÷ 3 = 5 exactly, but 9 is not divisible by 2 because 9÷ 2 is 4 with 1 left over.

Your task is to find the minimum sum of the array a that can be obtained by making as many transform operations as you want. Can you?

Input

The first line contains an integer T (1 ≤ T ≤ 100) specifying the number of test cases.

The first line of each test case contains an integer n (1 ≤ n ≤ 10⁵), in which n is the size of array a. Then a line follows containing n integers a₁, ..., a_n (1 ≤ a_i ≤ 10⁶), giving array a.

The sum of n overall test cases does not exceed 3 × 10⁶.

Output

For each test case, print a single line containing the minimum sum of the array a that can be obtained after making as many transform operations as you want.

Example

Input

1
5
2 2 3 6 6

Output

11

题解：有一个长度为n的数组，对于数组中的两个元素x,y如果满足y%x==0,则可以将y转换成x，求经过多次变换后数组的前n项和最小是多少。由于数据量很大，我们可以用map(去重），对容器中的每一个元素便历，对x的每一因子xi从小到大枚举，如果xi在容器中，则将x转换成xi,并结束循环，在求x的因子时如果将因子存进数组中在sort，会超时，所以我用了一个小技巧，闲话少说看代码。还需要注意的是mp.erase(),当在迭代器中调用这个函数的话，需要先返回上一个迭代器，不然指针会变为一个野指针（可参考链表理解），我在这个地方卡了好久哦。

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<stack>
 #include<iostream>
 #include<map>
 typedef long long ll;
 const int MAXN=+;
 const int INF=1e6;
 using namespace std;
 map<ll,ll>::iterator it;
 ll dp[];
 ll l=;
 int main()
 {
     ll m,n,t,i;
     //cout<<INF;
     scanf("%lld",&t);
     map<ll,ll>mp;
     while(t--)
     {
         mp.clear();
         scanf("%lld",&m);
         for(ll i=; i<m; i++)
         {
             scanf("%lld",&n);
             mp[n]++;
         }
         ll ans=,flag=;
         for(it=mp.begin(); it!=mp.end(); it++)
         {
             ll k=it->first;
             if(k==)
             {
                 flag=;
                 break;
             }
             ll flag=,ppq=;
             for(i=;i*i<=k;i++)
             {
                 if(k%i==)
                 {
                     if(mp.count(i))
                     {
                         mp[i]+=mp[k];
                         it--;
                         mp.erase(k);
                         flag=;
                         break;
                     }
                     else if(mp.count(k/i))
                     {
                         ppq=k/i;
                     }
                 }
             }
             if(!flag)
             {
                 if(ppq)
                 {
                     mp[ppq]+=mp[k];
                     it--;
                     mp.erase(k);
                 }
             }
         }
         if(flag)
             printf("%lld\n",m);
         else
         {
             for(it=mp.begin(); it!=mp.end(); it++)
             {
                 ans+=it->second*it->first;
             }
             cout<<ans<<endl;
         }
     }
     return ;
 }