Given a set of integers (it can contain equal elements).

You have to split it into two subsets A and B (both of them can contain equal elements or be empty). You have to maximize the value of mex(A)+mex(B).

Here mex of a set denotes the smallest non-negative integer that doesn't exist in the set. For example: 

  * mex(\{1,4,0,2,2,1\})=3 
  * mex(\{3,3,2,1,3,0,0\})=4 
  * mex(∅)=0 (mex for empty set) 



The set is splitted into two subsets A and B if for any integer number x the number of occurrences of x into this set is equal to the sum of the number of occurrences of x into A and the number of occurrences of x into B.

Input

The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer n (1≤ n≤ 100) — the size of the set.

The second line of each testcase contains n integers a_1,a_2,... a_n (0≤ a_i≤ 100) — the numbers in the set.

Output

For each test case, print the maximum value of mex(A)+mex(B).

Example

Input


4
6
0 2 1 5 0 1
3
0 1 2
4
0 2 0 1
6
1 2 3 4 5 6


Output


5
3
4
0

Note

In the first test case, A=\left\{0,1,2\right\},B=\left\{0,1,5\right\} is a possible choice.

In the second test case, A=\left\{0,1,2\right\},B=∅ is a possible choice.

In the third test case, A=\left\{0,1,2\right\},B=\left\{0\right\} is a possible choice.

In the fourth test case, A=\left\{1,3,5\right\},B=\left\{2,4,6\right\} is a possible choice.
1
100
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
 15
1
0
1
1
2
0 0
2
0 1
2
1 1
2
0 2
2
1 2
3
0 0 0
3
0 1 0
3
0 1 2
4
0 1 2 3
4
1 1 2 3
4
0 1 0 1
4
0 0 0 1
5
0 0 0 0 0

100
 1
0
2
2
0
1
0
2
3
3
4
0
4
3
2

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


public class p5 {

	/**

	1 2 3 4, 1 2 3 4 => 5, 5
	1 2 3, 1 2 3 4 => 4, 5
	2 3 4, 1 2 3 4 => 1, 5
	1 2 6 8 9, 1 2 3 4 => 1: 2, 2: 2, 3: 1, 4: 1 
	*/

	public static void doProb(int[] in) {
		int n = in.length;

		// just note where the count becomes 
		HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();

		// put counts in hashmap

		for (int i = 0; i < n; i++) {
			if (m.containsKey(in[i])) {
				int temp = m.get(in[i]);
				m.put(in[i], ++temp);
			} else {
				m.put(in[i], 1);
			}
		}

		// go through each key starting at bottom
		// and find where we go to 0

		int two_index_end = 0;
		boolean update_two_index_end = true;
		int one_index_end = 0;
		for (int i = 0; i < 100; i++) {
			if (!m.containsKey(i)) {
				one_index_end = i - 1;
				break;
			}

			if (m.get(i) == 1 && update_two_index_end) {
				two_index_end = i - 1;
				update_two_index_end = false;
			}
			// if (m.get(i) >= 2) {
			// 	continue;
			// }

			// so either we have 0, 1 


		}

		// System.out.println("two index end " + two_index_end);
		// System.out.println("one index end " + one_index_end);
		if (two_index_end < 0 && one_index_end < 0) {
			System.out.println("0");
		} else if (two_index_end < 0) {
			System.out.println(one_index_end + 1);
		} else {
			System.out.println(one_index_end + two_index_end + 2);
		}

		return;
	}

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);

		int t = scan.nextInt();
		for (int i = 0; i < t; i++) {
			int n = scan.nextInt();

			int[] input = new int[n];

			for (int j = 0; j < n; j++) {
				input[j] = scan.nextInt();
			}
			doProb(input);
		}
	}
}