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.util.*;
import java.io.*;

public class D {

    static class FastReader {
        StringTokenizer str;
        BufferedReader br;

        FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            if (str == null || !str.hasMoreTokens()) {
                try {
                    str = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                    return "";
                }
            }
            return str.nextToken();
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        int[] nextIntArray(int n) {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }

        long[] nextLongArray(int n) {
            long a[] = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
    }

    public static void main(String args[]) throws Exception {

        FastReader fr = new FastReader();
        int t = fr.nextInt();
        StringBuilder opt = new StringBuilder();

        while (t-- > 0) {

            int n = fr.nextInt();
            int a[] = new int[n + 1];
            for (int i = 0; i < n; i++)
                a[fr.nextInt()]++;

            Arrays.sort(a);

            if (n == 1) {
                opt.append("1\n");
                continue;
            }
            
            int ans = 0, ctr = a[n];

            for (int i = n; i >= 0 && a[i] != 0 && ctr > 0; i--) {
                if (a[i] >= ctr) {
                    ans += ctr;
                    ctr = a[i - 1] < ctr ? a[i - 1] : ctr - 1;
                }
            }

            opt.append(ans).append("\n");
        }

        System.out.print(opt);
    }
}