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 ll long long
#define pb push_back
#define all(x) x.begin(), x.end()
#define sortall(x) sort(all(x))
#define reverseall(x) reverse(all(x))
#define to_upper(x) transform(x.begin(), x.end(), x.begin(), ::toupper)
#define to_lower(x) transform(x.begin(), x.end(), x.begin(), ::tolower)
#define sz(x) (int)(x).size()
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define FOR(i,n) for(int i=0;i<n;i++)
#define FORS(i,s,n) for(int i=s;i<n;i++)
#define send {ios_base::sync_with_stdio(false);}
#define help {cin.tie(NULL); cout.tie(NULL);}
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "= " << x << ", " << #y << "= " << y << endl
#define clr(x) memset(x, 0, sizeof(x))
#define clrv(x, y) memset(x, y, sizeof(x))
#define gcd __gcd
#define PI 3.1415926535897932384626
typedef pair<int, int>	pii;
typedef pair<ll, ll>	pll;
typedef vector<int>		vi;
typedef vector<ll>		vl;
typedef vector<pii>		vpii;
typedef vector<pll>		vpll;
typedef vector<vi>		vvi;
typedef vector<vl>		vvl;
typedef priority_queue<int, vi, greater<int>> minpq;
typedef priority_queue<ll, vi, greater<ll>> minpql;

// DEBUG
void __print(int x) {cout << x;}
void __print(long x) {cout << x;}
void __print(long long x) {cout << x;}
void __print(unsigned x) {cout << x;}
void __print(unsigned long x) {cout << x;}
void __print(unsigned long long x) {cout << x;}
void __print(float x) {cout << x;}
void __print(double x) {cout << x;}
void __print(long double x) {cout << x;}
void __print(char x) {cout << '\'' << x << '\'';}
void __print(const char *x) {cout << '\"' << x << '\"';}
void __print(const string &x) {cout << '\"' << x << '\"';}
void __print(bool x) {cout << (x ? "true" : "false");}

template <typename T, typename V>
void __print(const pair<T, V> &x);
template <typename T>
void __print(const vector<T> &x);

template<typename T, typename V>
void __print(const pair<T, V> &x) {cout << '{'; __print(x.first); cout << ','; __print(x.second); cout << "}";}
template<typename T>
void __print(const vector<T> &x) {cout << "{";for(int u69=0;u69<x.size();u69++){__print(x[u69]);cout << (u69+1==x.size()?"":",");};cout << "}";}
template<typename T>
void __print(const T &x) {int f = 0; cout << '{'; for (auto &i: x) cout << (f++ ? "," : ""), __print(i); cout << "}";}
void _print() {cout << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cout << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cout << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
//
const int mod = 1000000007;
const double eps = 1e-14;
void YES(bool t = 1) { const string YESNO[2] = {"NO", "YES"};cout << YESNO[t] << endl; }
void Yes(bool t = 1) { const string YesNo[2] = {"No", "Yes"};cout << YesNo[t] << endl; }
void yes(bool t = 1) { const string yesno[2] = {"no", "yes"};cout << yesno[t] << endl; }

struct custom_hash {    //Custom Hash_func for unordered_map
    /*OR DO
        mp.max_load_factor(0.25);
        mp.reserve(1024);
    */
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};
ll mpow(int b, int e, int modu=0, int mod=1000000007) {
    ll result = 1,base = b, exp = e;
    if(modu) base %= mod;
    while (exp > 0) {
        if (exp & 1){
            result = ((ll)result * base) ;
            if(modu) result %= mod;
        }
        base = ((ll)base * base);
        if(modu) base %= mod;
        exp >>= 1;
    }
    return result;
}
inline bool isPowerOfTwo(ll x) { return x && (!(x&(x-1))); } 
bool isPerfectSquare(long double x){long double sr = sqrt(x); return ((sr - floor(sr)) == 0);} 
bool isIn(string &s2, string &s1){if (s1.find(s2) != string::npos) return true;return false;}
bool isSorted(vi &arr){for(int i=0;i<(int)arr.size()-1;i++) if(arr[i] > arr[i+1]) return false;return true;}
/* stuff you should look for
    * int overflow, array bounds
    * special cases (n=1?)
    * do smth instead of nothing and stay organized
    * WRITE STUFF DOWN
    * DON'T GET STUCK ON ONE APPROACH
*/

int TEST_CASES = 1;
void submain(){
    int n;
    cin >> n;
    vl mp[n+1];
    vl arr(n), brr(n);
    FOR(i,n){
        cin >> arr[i];
    }
    FOR(i, n){
        cin >> brr[i];
    }
    FOR(i,n){
        mp[arr[i]].pb(brr[i]);   
    }
    
    vl ans(n+1);
    for(int j=1;j<=n;j++){
        sortall(mp[j]);
        reverseall(mp[j]);
        for(int i=1;i<mp[j].size();i++){
            mp[j][i] += mp[j][i-1];
        }
        for(int k=1;k<=mp[j].size();k++){
            ll modd = mp[j].size()%k;
            // debug(mp[j]);
            ans[k] += (ll)mp[j][mp[j].size() - modd - 1];
        }
    }
    
    FORS(i,1,n+1){
        cout << ans[i] << ' ';
    }
    cout << '\n';
    
}


int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    #endif
    send help
    //// Preprocessing space

    ////
    if(TEST_CASES){
        int t;
        cin >> t;
        while(t--){
            submain();
        }
    }else{
        submain();
    }
    return 0;    
}