Problem: Polycarp is an organizer of a Berland ICPC regional event. There are n universities in Berland numbered from 1 to n. Polycarp knows all competitive programmers in the region. There are n students: the i-th student is enrolled at a university u_i and has a programming skill s_i.

Polycarp has to decide on the rules now. In particular, the number of members in the team.

Polycarp knows that if he chooses the size of the team to be some integer k, each university will send their k strongest (with the highest programming skill s) students in the first team, the next k strongest students in the second team and so on. If there are fewer than k students left, then the team can't be formed. Note that there might be universities that send zero teams.

The strength of the region is the total skill of the members of all present teams. If there are no teams present, then the strength is 0.

Help Polycarp to find the strength of the region for each choice of k from 1 to n.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

The first line of each testcase contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of universities and the number of students.

The second line of each testcase contains n integers u_1, u_2, ..., u_n (1 ≤ u_i ≤ n) — the university the i-th student is enrolled at.

The third line of each testcase contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) — the programming skill of the i-th student.

The sum of n over all testcases doesn't exceed 2 ⋅ 10^5.

Output

For each testcase print n integers: the strength of the region — the total skill of the members of the present teams — for each choice of team size k.

Example

Input


4
7
1 2 1 2 1 2 1
6 8 3 1 5 1 5
10
1 1 1 2 2 2 2 3 3 3
3435 3014 2241 2233 2893 2102 2286 2175 1961 2567
6
3 3 3 3 3 3
5 9 6 7 9 7
1
1
3083


Output


29 28 26 19 0 0 0 
24907 20705 22805 9514 0 0 0 0 0 0 
43 43 43 32 38 43 
3083 

Note

In the first testcase the teams from each university for each k are: 

  * k=1: 
    * university 1: [6], [5], [5], [3]; 
    * university 2: [8], [1], [1]; 
  * k=2: 
    * university 1: [6, 5], [5, 3]; 
    * university 2: [8, 1]; 
  * k=3: 
    * university 1: [6, 5, 5]; 
    * university 2: [8, 1, 1]; 
  * k=4: 
    * university 1: [6, 5, 5, 3]; 

Solution (in CPP):

Solution:
///*.......................Shadman Sakib Utshob............................*///

#include <bits/stdc++.h>
using namespace std;

//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;

#define Boost           ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0)
#define see(x)        cerr<< __LINE__ << ": " << (#x) << " is "<< (x) <<'\n';
#define pb              push_back
#define inf             1e19
#define rep(i,n)        for(int i=0;i<n;i++)
#define rep1(i,n)       for(int i=1;i<=n;i++)
#define ll              long long
#define MOD             1000000007
#define mem(x,i)        memset(x,i,sizeof x)
#define Ones(x)         __builtin_popcount(x);
#define PI              acos(-1.0)
#define ff              first
#define ss              second
#define T               int t;cin>>t;while(t--)
#define ok              cout<<"ok"<<endl;
#define all(x)          (x).begin(),(x).end()
#define ordered_set     tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define nln	            cout<<endl;
#define yes	      		cout<<"YES"<<endl;
#define no 	      		cout<<"NO"<<endl;
#define maxn 300005

int dx[]= {-1,0,1,0}; //up,right,down,left
int dy[]= {0,1,0,-1};

int main()
{
    Boost;

    T
    {
        ll n,i,j,k,x,sum=0,mx=-1;
        cin>>n;
        ll u[n+2],s[n+2],ans[n+2];
        mem(ans,0);
        vector<ll>pre[n+2],v[n+2],kaj;

        rep(i,n)
        {
            cin>>u[i];
        }
        rep(i,n)
        {
            cin>>s[i];
            v[u[i]].pb(s[i]);
        }
        for(i=1; i<=n; i++)
        {
            if(v[i].size())
                kaj.pb(i);
        }
        for(i=0; i<kaj.size(); i++)
        {
            sort(v[kaj[i]].begin(),v[kaj[i]].end(),greater<ll>());

        }

        for(i=0; i<kaj.size(); i++)
        {
            for(j=0; j<v[kaj[i]].size(); j++)
            {
                if(j==0)
                {
                    pre[kaj[i]].pb(v[kaj[i]][0]);
                }
                else
                {
                    pre[kaj[i]].pb(pre[kaj[i]][j-1]+v[kaj[i]][j]);
                }
            }
        }

        for(i=0; i<kaj.size(); i++)
        {
            x=kaj[i];//index
            for(j=1; j<=v[x].size(); j++)
            {
                ll len=v[x].size();
                if(len%j==0)
                {
                    ans[j]+=pre[x][pre[x].size()-1];
                }
                else
                {
                    ll r=len%j;
                    ll y = len - r - 1;
                    ans[j]+=pre[x][y];
                }
            }
        }

        for(i=1;i<=n;i++)
        {
            cout<<ans[i]<<" ";
        }nln

    }




}
/*
1
5
1 1 1 1 1
4 4 4 4 4

*/