This problem is actually a subproblem of problem G from the same contest.
There are n candies in a candy box. The type of the i-th candy is ai (1≤ai≤n).
You have to prepare a gift using some of these candies with the following restriction: the numbers of candies of each type presented in a gift should be all distinct (i. e. for example, a gift having two candies of type 1 and two candies of type 2 is bad).
It is possible that multiple types of candies are completely absent from the gift. It is also possible that not all candies of some types will be taken to a gift.
Your task is to find out the maximum possible size of the single gift you can prepare using the candies you have.
You have to answer q independent queries.
If you are Python programmer, consider using PyPy instead of Python when you submit your code.
Input
The first line of the input contains one integer q (1≤q≤2⋅10^5) — the number of queries. Each query is represented by two lines.
The first line of each query contains one integer n (1≤n≤2⋅10^5) — the number of candies.
The second line of each query contains n integers a1,a2,…,an (1≤ai≤n), where ai is the type of the i-th candy in the box.
It is guaranteed that the sum of n over all queries does not exceed 2⋅10^5.
Output
For each query print one integer — the maximum possible size of the single gift you can compose using candies you got in this query with the restriction described in the problem statement.
Example
Input
3
8
1 4 8 4 5 6 3 8
16
2 1 3 3 4 3 4 4 1 3 2 2 2 4 1 1
9
2 2 4 4 4 7 7 7 7
Output
3
10
9
Note
In the first query, you can prepare a gift with two candies of type 8 and one candy of type 5, totalling to 3 candies.
Note that this is not the only possible solution — taking two candies of type 4 and one candy of type 6 is also valid.

题意：给定n个糖果，每个糖果有一个种类，现在要挑出一堆尽可能多的糖果，这些糖果中，不同种类的糖果的个数必须互不相同，求这堆糖果最大有多少。
统计出不同种类的糖果个数，对其排序，从最大的开始选。
贪心地想，最大的一定计入答案。
设上一次计入答案的糖果数为k，当前枚举到的糖果数为s[i]，则当次计入到答案的糖果数为ans=min(k-1,s[i])

#include<bits/stdc++.h>
using namespace std;
int vis[200010];
int visans[200010],t;
queue<int> cnt;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int q,n,a;
    cin>>q;
    while(q--)
    {
        int ans=0;int b;t=0;
        cin>>n;
        for(int i=1;i<=n;++i)
        {
            cin>>a;
            if(!vis[a]) cnt.push(a);
            ++vis[a];
        }
        for(int i=1;i<=n;++i)
        {
            if(vis[i])
            {
                visans[++t]=vis[i];
            }
        }
        sort(visans+1,visans+1+t);
        //for(int i=1;i<=t;++i) cout<<visans[i]<<' ';
        int k=visans[t];ans+=k;
        for(int i=t-1;i;--i)
        {
            if(k<1) break;
            if(k>visans[i]) k=visans[i];
            else --k;
            ans+=k;
        }
        cout<<ans<<endl;
        while(!cnt.empty())
        {
            vis[cnt.front()]=0;
            cnt.pop();
        }

    }
    return 0;
}