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 Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter wr = new PrintWriter(System.out);
        int q = Integer.parseInt(br.readLine().trim());
        while(q-->0){
            int N = Integer.parseInt(br.readLine().trim());
            String[] S = br.readLine().split(" ");
            int[] A = new int[N];
            Map<Integer,Integer> mp = new HashMap<Integer,Integer>();
            for (int i=0;i<N;i++) {
                A[i] = Integer.parseInt(S[i]);
                if(!mp.containsKey(A[i])){
                    mp.put(A[i],0);
                }
                mp.put(A[i],mp.get(A[i])+1);
            }
            Integer v[] = new Integer[mp.size()];
            v  = (Integer[])mp.values().toArray(v);
            Arrays.sort(v);
            int ans = v[v.length-1];
            int last = ans-1;
            for(int i=v.length-2;i>=0;i--){
                if(last==0){
                    break;
                }
                int taken = Math.min(last,v[i]);
                ans = ans + taken;
                last = taken-1;
            }
            wr.println(ans);
        }

        wr.flush();
        wr.close();
    }

    static long min_diff(long[] A) {
        // Write your code here
        Arrays.sort(A);
        long ans = Long.MAX_VALUE;
        for (int i = 0; i < A.length - 1; i++) {
            long x = A[i];
            if (i - 1 >= 0) {
                ans = Math.min(ans, Math.abs(A[i] - A[i - 1]));
            }
            if (i + 1 < A.length ) {
                ans = Math.min(ans, Math.abs(A[i] - A[i + 1]));
            }
            if (ans == 0) {
                break;
            }
            if (A[i] % 2 == 0) {
                x /= 2;
            } else {
                x *= 2;
            }
            int pos = i;
            boolean found = false;
            if(A[i]%2==0 && i-1>=0) {
                pos = Arrays.binarySearch(A, 0,i-1, x);
                found = true;
            }
            else if(A[i]%2==1 && i+1<=A.length-1){
                pos = Arrays.binarySearch(A,i+1,A.length-1,x);
                found = true;
            }
            if(!found){
                continue;
            }

            if (pos < 0) {
                pos = -pos - 1;
                ans = Math.min(ans, Math.abs(A[pos] - x));
                if (pos - 1 != i) {
                    ans = Math.min(ans, Math.abs(A[pos - 1] - x));
                }
            } else {
                ans = 0;
                break;
            }


        }
        return ans;

    }
}