You are given an array a consisting of n integers a1, ..., an. In one operation, you can choose 2 elements ai and aj in which ai is divisible by aj and transform ai to aj.
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, ..., an (1 ≤ ai ≤ 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 athat can be obtained after making as many transform operations as you want.
Example
Input
1 5 2 2 3 6 6
Output
11
废柴的一天 。。。。
这题用了埃氏筛法，1e6*1e6的复杂度过了 我也是醉了
思维：用一个数组V记录每个数字出现的次数 模拟埃氏筛 如果某个 I 被筛到 I*J那么 吧 J 的数字全部变成I
WA因 ：因为long long 是lld输出 （输出格式错误）
#include<bits/stdc++.h>
using namespace std;
int t,n,a[100005],v[1000005];
int main(){
scanf("%d",&t);
while(t--){
long long sum=0;
scanf("%d",&n);
memset(v,0,sizeof(v));
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
v[a[i]]++;
sum+=a[i];
}
for(int i=1;i<=1000000;i++){
if(v[i]){
for(int j=2;i*j<=1000000;j++)
if(v[i*j]){
sum=sum-v[i*j]*i*j+i*v[i*j];
v[j*i]=0;
}
}
}
printf("%lld\n",sum);//。。有点崩溃 这是输出格式错误的原因
}
return 0;
}