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.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.TreeMap;
import java.util.Map;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Xinyi Tao
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskDR570D3 solver = new TaskDR570D3();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskDR570D3 {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int q = in.nextInt();
            for (int i = 0; i < q; i++) {
                int n = in.nextInt();
                TreeMap<Integer, Integer> m = new TreeMap<>();
                TreeMap<Integer, Integer> r = new TreeMap<>();
                for (int j = 0; j < n; j++) {
                    int ai = in.nextInt();
                    m.merge(ai, 1, Integer::sum);
                }
                for (Integer j : m.values()) {
                    r.merge(j, 1, Integer::sum);
                }
                Integer p = null;
                Integer s = 0;
                for (Integer j : r.descendingKeySet()) {
                    Integer v = r.get(j);
                    Integer a0;
                    Integer am;
                    if (p != null)
                        am = Math.min(p - 1, j);
                    else
                        am = j;
                    a0 = Math.max(am - v + 1, 1);
                    if (am < a0) break;
                    if (a0 < 1) break;
                    s += (a0 + am) * (am - a0 + 1) / 2;
                    p = a0;
                    if (p == 1) break;
                }

                out.println(s);
            }
        }

    }
}