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:
#include<bits/stdc++.h>
using namespace std;
#define T top()
#define E end()
#define PS push
#define Q queue
#define PP pop()
#define L0 (LL)0
#define V vector
#define ST stack
#define FI first
#define MAX 4e18
#define MIN -MAX
#define DQ deque
#define SZ size()
#define C clear()
#define B begin()
#define F front()
#define SE second
#define DL double
#define G getline
#define IN insert
#define endl "\n"
#define EM empty()
#define cnn continue
#define MD 1000000007
#define PSB push_back
#define PI acos(-1.00)
#define PPB pop_back()
#define PSF push_front
#define ub upper_bound
#define lb lower_bound
#define PPF pop_front()
#define CF cout.flush()
#define line cout<<endl
#define LL long long int
#define one cout<<1<<endl
#define zer cout<<0<<endl
#define PQ priority_queue
#define nil cout<<-1<<endl
#define N cout<<"NO"<<endl
#define NN cout<<"No"<<endl
#define Y cout<<"YES"<<endl
#define YY cout<<"Yes"<<endl
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#define MAU cout<<"Case "<<mau<<": "
#define SP(pre) fixed<<setprecision(pre)
#pragma GCC optimization ("unroll-loops")
#define MMAU cout<<"Case "<<mau<<":"<<endl
#define __lcm(A1,A2) (A1/(__gcd(A1,A2)))*A2
#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define file freopen("input.txt","r",stdin);freopen("output.txt","w",stdout)
#define UOSMOY v.C;w.C;ww.C;uu.C;uo.C;vo.C;vu.C;ab.C;bc.C;cc.C;uuu.C;a.C;b.C;c.C;u.C
V<LL>v;
V<LL>w,ab,bc;
set<LL>uo;
map<LL,LL>ww,cc,u;
V<pair<LL,LL> >uu,su;
map<LL,V<LL> >uuu,suu;
map<pair<LL,LL>,LL> vo;
map<pair<LL,LL>,LL> vu;
map<LL,pair<LL,pair<LL,LL > > > vs;
priority_queue<LL,V<LL>,greater<LL> > moy;
LL dx[]= {-1,0,1,0,1,1,-1,-1};
LL dy[]= {0,1,0,-1,-1,1,1,-1};
LL dxx[]= {2,1,-1,-2,-2,-1,1,2};
LL dyy[]= {1,2,2,1,-1,-2,-2,-1};
LL dp[101][10001];
LL ar[10][10];
int main()
{
//  BRR;
//  file;
    FAST;
    char ch;
    string a,b,c;
    DL x,y,d,e,f,g,h;
    LL n,i,j,k,p,q,o,l,s,tt=1,m,t,z,aa,r=0;
    cin>>tt;
    for(LL mau=1; mau<=tt; mau++)
    {
        UOSMOY;
        cin>>n;
        for(i=0; i<n; i++)
        {
            cin>>o;
            v.PSB(o);
        }
        for(i=0; i<n; i++)
        {
            cin>>o;
            uuu[v[i]].PSB(o);
        }
        suu.C;
        for(auto it=uuu.B; it!=uuu.E; it++)
        {
            sort(it->SE.B,it->SE.E,greater<LL>());
            k=0;
            o=it->SE.SZ;
            for(i=0; i<o; i++)
            {
                k+=it->SE[i];
                suu[o*1000000+(it->FI)].PSB(k);
            }
        }
        for(i=1; i<=n; i++)
        {
            l=0;
            for(auto it=suu.rbegin(); it!=suu.rend(); it++)
            {
                o=it->SE.SZ;
                if(o<i)
                {
                    break;
                }
                o/=i;
                o*=i;
                o--;
                l+=(it->SE)[o];
            }
            cout<<l<<" ";
        }
        line;
    }
    return 0;
    //IN
    //AL
}