Educational Codeforces Round 108 (Rated for Div. 2) C. Berland Regional（模拟）

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

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 𝑘kstudents 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 𝑛n (1≤n≤2⋅10^5) — the number of universities and the number of students.

The second line of each testcase contains 𝑛n integers u1,u2,…,un (1≤ui≤n) — the university the 𝑖i-th student is enrolled at.

The third line of each testcase contains 𝑛n integers s1,s2,…,sn (1≤si≤10^9) — the programming skill of the 𝑖i-th student.

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

Output

For each testcase print 𝑛n 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

Copy

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

Copy

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

直接按照题意模拟，首先开vector数组用来维护每个大学的学生，可以直接添加完后sort，将每个大学的学生按照skill降序排列。然后遍历出现过的大学（否则会T），先对排序后的学生的skill做一次前缀和，然后计算这个大学对于k = 1 to n的贡献。pos = v[uni[i]].size() / k * k;表示需要的前缀和的位置。注意如果pos == 0说明对于以后的k这个大学都凑不满一组了，此时必须直接break，否则会T。

#include <bits/stdc++.h>
#define int long long
using namespace std;
int n, u[200005], s[200005];
int ans[200005];
bool vis[200005];
void primes(int n) {
	memset(vis, 0,sizeof(vis));
	for(int i = 2; i <= n; i++) {
		if(vis[i]) continue;
		p.push_back(i);
		for(int j = i; i * j <= n; j++) vis[i * j] = 1;
	}
}
bool cmp(int a, int b) {
	return a > b;
}
signed main() {
	int t;
	cin >> t;
	while(t--) {
		scanf("%lld", &n);
		vector<int> v[200005];
		for(int i = 1; i <= n; i++) ans[i] = 0;
		ans[0] = 0;
		bool b[200005];
		vector<int> uni;
		for(int i = 1; i <= n; i++) b[i] = 0;
		for(int i = 1; i <= n; i++) {scanf("%d", &u[i]); b[u[i]] = 1;}
		for(int i = 1; i <= n; i++) scanf("%d", &s[i]);
		for(int i = 1; i <= n; i++) {
			v[u[i]].push_back(s[i]);
		}
		for(int i = 1; i <= n; i++) {
			if(b[i]) uni.push_back(i);
		}
		for(int i = 1; i <= n; i++) {
			sort(v[i].begin(), v[i].end(), cmp);
		}
		for(int i = 0; i < uni.size(); i++) {//只需要遍历出现的大学
			long long sum = 0;
			vector<long long> pre;//前缀和vector
			pre.push_back(0);
			for(int j = 0; j < v[uni[i]].size(); j++) {
				sum += 1ll * v[uni[i]][j];
				pre.push_back(sum);
			}
			for(int k = 1; k <= n; k++) {//遍历可能的分组计算贡献
				int pos = v[uni[i]].size() / k * k;
				if(pos == 0) break;//注意及时break 这样能保证运行时间不爆炸
				if(pos < pre.size()) ans[k] += pre[pos];
			}
		}
 
		for(int i = 1; i <= n; i++) cout << ans[i] << " ";
		cout << endl;
	}
	return 0;
}