Problem: 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 a_i (1 ≤ a_i ≤ 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 a_1, a_2, ..., a_n (1 ≤ a_i ≤ n), where a_i 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.

Solution (in JAVA):

Solution:
//package solution;

import java.io.*;
import java.math.BigInteger;
import java.util.*;

public final class Solution {
    BufferedReader br;
    StringTokenizer stk;

    public static void main(String[] args) throws Exception {
        new Solution().run();
    }
    
    {
        stk = null;
        br = new BufferedReader(new InputStreamReader(System.in));
    }
    
    long mod = 1000_000_000 + 7;
    StringBuilder res = new StringBuilder(1000005);
    
    void run() throws Exception {
        int q = ni();
        while(q-- > 0) {
            int n = ni();
            TreeSet<Integer> available = returnSet(n);
            int[] a = new int[n + 1];
            for(int i = 0; i < n; i++) {
                int num = ni();
                a[num]++;
            }
            
            int count = 0;
            for(int i = n; i > 0; i--) {
                if(a[i] == 0) {
                    continue;
                }
                Integer avail = available.floor(a[i]);
                if(avail != null) {
                    available.remove(avail);
                    count += avail;
                }
            }
            
            res.append(count).append("\n");
        }
        System.out.print(res);
    }
    
    TreeSet<Integer> returnSet(int n) {
        TreeSet<Integer> set = new TreeSet<>();
        for(int i = 1; i <= n; i++) {
            set.add(i);
        }
        return set;
    }
    
    //Reader & Writer
    String nextToken() throws Exception {
        if (stk == null || !stk.hasMoreTokens()) {
            stk = new StringTokenizer(br.readLine(), " ");
        }
        return stk.nextToken();
    }

    String nt() throws Exception {
        return nextToken();
    }

    int ni() throws Exception {
        return Integer.parseInt(nextToken());
    }

    long nl() throws Exception {
        return Long.parseLong(nextToken());
    }
}