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 CF1183G extends PrintWriter {
	CF1183G() { super(System.out); }
	static class Scanner_ {
		int k, l; byte[] bb = new byte[1 << 15];
		byte getc() {
			if (k >= l) {
				try { l = System.in.read(bb); } catch (IOException e) { return -1; }
				if (l <= 0) return -1; k = 0;
			}
			return bb[k++];
		}
		int nextInt() {
			byte c = 0; while (c <= 32) c = getc();
			boolean minus = c == '-'; if (minus) c = getc();
			int a = 0; while (c > 32) { a = a * 10 + c - '0'; c = getc(); }
			return minus ? -a : a;
		}
	}
	Scanner_ sc = new Scanner_();
	public static void main(String[] __) { new CF1183G().main(); }

	void main() {
		int q = sc.nextInt();
		while (q-- > 0) {
			int n = sc.nextInt();
			int[] kk = new int[n];
			int[] ll = new int[n];
			for (int i = 0; i < n; i++) {
				int a = sc.nextInt() - 1;
				int f = sc.nextInt();
				kk[a]++;
				if (f == 1)
					ll[a]++;
			}
			Integer[] ii = new Integer[n];
			for (int i = 0; i < n; i++)
				ii[i] = i;
			Arrays.sort(ii, (i, j) -> kk[i] != kk[j] ? kk[j] - kk[i] : ll[j] - ll[i]);
			TreeMap<Integer, Integer> mp = new TreeMap<>();
			int k_ = 0, l_ = 0;
			for (int k = n, h = 0, i; k > 0 && h < n; k--) {
				while (h < n && kk[i = ii[h]] >= k) {
					mp.put(ll[i], mp.getOrDefault(ll[i], 0) + 1);
					h++;
				}
				if (mp.isEmpty())
					continue;
				Map.Entry<Integer, Integer> entry = mp.lastEntry();
				int l = entry.getKey();
				int cnt = entry.getValue();
				if (cnt == 1)
					mp.remove(l);
				else
					mp.put(l, cnt - 1);
				k_ += k;
				l_ += Math.min(k, l);
			}
			println(k_ + " " + l_);
		}
		flush();
	}
}