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Problem Description Every school has some legends, Northeastern University is the same.

Enter from the north gate of Northeastern University，You are facing the main building of Northeastern University.Ninety-nine percent of the students have not been there，It is said that there is a monster in it.

QSCI am a curious NEU_ACMer，This is the story he told us.

It’s a certain period，QSCI am in a dark night, secretly sneaked into the East Building，hope to see the master.After a serious search，He finally saw the little master in a dark corner. The master said：

“You and I, we’re interfacing.please solve my little puzzle！

There are N pairs of numbers，Each pair consists of a key and a value，Now you need to move out some of the pairs to get the score.You can move out two continuous pairs，if and only if their keys are non coprime(their gcd is not one).The final score you get is the sum of all pair’s value which be moved out. May I ask how many points you can get the most？

The answer you give is directly related to your final exam results~The young man~”

QSC is very sad when he told the story，He failed his linear algebra that year because he didn’t work out the puzzle.

Could you solve this puzzle?

（Data range：1<=N<=300

1<=Ai.key<=1,000,000,000 0<Ai.value<=1,000,000,000）

Input

First line contains a integer T，means there are T(1≤T≤10) test case。

Each test case start with one integer N . Next line contains N integers，means Ai.key.Next line contains N integers，means Ai.value.

Output

For each test case,output the max score you could get in a line.

Sample Input

3 3 1 2 3 1 1 1 3 1 2 4 1 1 1 4 1 3 4 3 1 1 1 1

Sample Output

0 2 0

Source

2016 ACM/ICPC Asia Regional Shenyang Online

题意：给n对数,如果相邻的一对数的第一个的gcd！=1,那么这两个数就可以一块拿走,获得第二个数的和的收益,求最大的收益。

思路：这里有点类似于求回文子串的做法，求回文子串的时候是判断dp【l】【r】能否组成回文，这里我们dp【l】【r】为区间【l，r】的最大收益，如果dp【l+1】【r-1】等于sum【r-1】-sum【l】就代表区间【l+1，r-1】的gcd大于1，如此一来只要gcd（a【i】，a【j】）再大于1的话就可以直接合并了。

#include
   
    using namespace std;typedef long long ll;const int maxn=305;ll sum[maxn],a[maxn],dp[maxn][maxn],v[maxn];ll dfs(int l,int r){       if(dp[l][r]!=-1) return dp[l][r];    ll ans=-1;    if(l==r) ans=0;    else if(l+1==r) {           if(__gcd(a[l],a[r])>1) ans=v[l]+v[r];        else ans=0;    }    else {           ll temp=__gcd(a[l],a[r]);        if(temp>1&&dfs(l+1,r-1)==sum[r-1]-sum[l]) ans=sum[r]-sum[l-1];        for(int k=l;k
   

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