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.*;
import java.util.*;

public class Codeforces1183 {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(
                new InputStreamReader(System.in));

        StringTokenizer st = new StringTokenizer(br.readLine());
        int q = Integer.parseInt(st.nextToken());
        int[][] results = new int[q][];
        for (int i = 0; i < q; i++) {
            st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            int[][] candies = new int[n][];

            for (int j = 0; j < n; j++) {
                st = new StringTokenizer(br.readLine());
                candies[j] = new int[]{Integer.parseInt(st.nextToken()), Integer.parseInt(st.nextToken())};
            }

            int[] result = packTheBox(candies);
            results[i] = result;
        }


        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out, "UTF-8"), 512));
        for (int i = 0; i < q; i++) {
            out.print(results[i][0]);
            out.print(" ");
            out.print(results[i][1]);
            out.println();
        }
        out.flush();
    }

    private static int[] packTheBox(int[][] candies) {
        Map<Integer, int[]> candyToCount = new HashMap<>();
        for (int i = 0; i < candies.length; i++) {
            int[] count = candyToCount.get(candies[i][0]);
            if (count == null) {
                count = new int[2];
                candyToCount.put(candies[i][0], count);
            }
            count[0] += 1;
            count[1] += candies[i][1];
        }

        List<int[]> counts = new ArrayList<>(candyToCount.values());
        Collections.sort(counts, (a, b) -> Integer.compare(b[0], a[0]));
        Queue<int[]> queue = new PriorityQueue<>(Comparator.comparingInt(a -> -a[1]));
        int maxCandies = 0;
        int maxOnes = 0;
        int maxUniqueCount = Integer.MAX_VALUE;

        for (int i = 0; i < counts.size(); i++) {
            int[] count = counts.get(i);

            while (maxUniqueCount > count[0] && !queue.isEmpty()) {
                int[] peek = queue.poll();
                maxCandies += maxUniqueCount;
                maxOnes += Math.min(peek[1], maxUniqueCount);
                maxUniqueCount--;
            }

            if (queue.isEmpty()) {
                maxUniqueCount = count[0];
            }
            queue.add(count);
        }

        while (!queue.isEmpty() && maxUniqueCount > 0) {
            int[] peek = queue.poll();
            maxCandies += maxUniqueCount;
            maxOnes += Math.min(peek[1], maxUniqueCount);
            maxUniqueCount--;
        }

        return new int[]{maxCandies, maxOnes};
    }
}