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:
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.Map;

/**
 * Built using CHelper plug-in Actual solution is at the top
 */
public class Practice {

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskA {

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int q = in.nextInt();
            for (int k = 0; k < q; k++) {
                int n = in.nextInt();
                HashMap<Integer, Integer> map = new HashMap<>();
                int maxFreq = 1;
                for (int i = 0; i < n; i++) {
                    int num = in.nextInt();
                    if (map.containsKey(num)) {
                        map.put(num, map.get(num) + 1);
                        maxFreq = Math.max(maxFreq, map.get(num));
                    } else {
                        map.put(num, 1);
                    }
                }
                long ans = 0;
                LinkedHashMap<Integer, Integer> reverseSortedMap = new LinkedHashMap<>();
                map.entrySet()
                        .stream()
                        .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
                        .forEachOrdered(x -> reverseSortedMap.put(x.getKey(), x.getValue()));
                for (Integer key : reverseSortedMap.keySet()) {
                    if (reverseSortedMap.get(key) >= maxFreq) {
                        ans += maxFreq;
                        maxFreq--;
                        
                    }
                    else{
                        ans+=reverseSortedMap.get(key);
                        maxFreq=reverseSortedMap.get(key)-1;
                    }
                    if(maxFreq==0){
                            break;
                        }
                }
                out.println(ans);
            }
        }
    }

    public long GCD(long a, long b) {
        if (b == 0) {
            return a;
        }
        return GCD(b, a % b);
    }

    public static long nCr(int n, int r) {
        return n * (n - 1) / 2;
    }

    static class InputReader {

        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

    }
}