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=1000+10;
const int INF=1e6;
using namespace std;
map<ll,ll>::iterator it;
ll dp[1000];
ll l=0;
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=0; i<m; i++)
        {
            scanf("%lld",&n);
            mp[n]++;
        }
        ll ans=0,flag=0;
        for(it=mp.begin(); it!=mp.end(); it++)
        {
            ll k=it->first;
            if(k==1)
            {
                flag=1;
                break;
            }
            ll flag=0,ppq=0;
            for(i=2;i*i<=k;i++)
            {
                if(k%i==0)
                {
                    if(mp.count(i))
                    {
                        mp[i]+=mp[k];
                        it--;
                        mp.erase(k);
                        flag=1;
                        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
",m);
        else
        {
            for(it=mp.begin(); it!=mp.end(); it++)
            {
                ans+=it->second*it->first;
            }
            cout<<ans<<endl;
        }
    }
    return 0;
}