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 CandyBox {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int t = Integer.parseInt(st.nextToken());
        for (int i = 0; i < t; i++) {
            int n = Integer.parseInt(br.readLine());
            HashMap<Integer, Integer> map = new HashMap<>();
            st = new StringTokenizer(br.readLine());
            for (int j = 0; j < n; j++) {
                int x = Integer.parseInt(st.nextToken());
                if (map.containsKey(x)) {
                    map.put(x, map.get(x) + 1);
                } else {
                    map.put(x, 1);
                }
            }

            TreeMap<Integer, Integer> valuecounts = new TreeMap<>(Collections.reverseOrder());
            for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
                if (valuecounts.containsKey(entry.getValue())) {
                    valuecounts.put(entry.getValue(), valuecounts.get(entry.getValue()) + 1);
                } else {
                    valuecounts.put(entry.getValue(), 1);
                }
            }

            int curr = Integer.MAX_VALUE;
            int ans = 0;
            for (Map.Entry<Integer, Integer> entry : valuecounts.entrySet()) {
                curr = Math.min(curr, entry.getKey());

                int num = Math.min(curr, entry.getValue());
                int l = curr;
                int f = Math.max(1, curr - entry.getValue() + 1);
                ans += num * (l + f) / 2;
                curr -= Math.min(curr, entry.getValue());
                if (curr <= 0) break;

            }

            System.out.println(ans);
        }
    }
}