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

#include <bits/stdc++.h>
using namespace std;
inline bool compare(long long long int x, long long long int y) {
  return x > y;
}
template <typename T>
T binaryExpo(T n, T p) {
  if (p == 0) {
    return 1LL;
  }
  if (p & 1) {
    return n * binaryExpo(n, p - 1);
  }
  T ret = binaryExpo(n, p / 2);
  return ret * ret;
}
template <typename T>
T bigMod(T n, T p, T m) {
  if (p == 0) {
    return 1LL % m;
  }
  if (p & 1) {
    return (n % m * bigMod(n, p - 1, m)) % m;
  }
  T ret = bigMod(n, p / 2, m) % m;
  return (ret * ret) % m;
}
template <typename T>
T modInv(T n, T m) {
  return bigMod(n, m - 2, m);
}
const long long long int maxN = 2e5 + 10;
const long long long int maxB = 320;
const long long long int INF = 9187201950435737471;
const long int INF_I = 2139062143;
const long long long int maxA = 1e6 + 10;
const long long long int MOD = 1e9 + 7;
const long long long int mINF = -9187201950435737472;
const long int mINF_I = -2139062144;
bool FILE_PRINT = 0;
long int marked[+2];
vector<long long long int> prime;
void sieve(long long long int n);
bool is_prime(long long long int num);
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(15);
  cout << fixed;
  long long long int number_of_test_cases = 1, test_case;
  cin >> number_of_test_cases;
  for (test_case = 1; test_case <= number_of_test_cases; test_case++) {
    long long long int i, j, k, m, n, query_type, q;
    cin >> n;
    long long long int arr[n + 10];
    for (i = (0); i < (n); ++i) cin >> arr[i];
    map<long long long int, long long long int> mp;
    for (i = 0; i < n; i++) {
      mp[arr[i]]++;
    }
    long long long int a = INF, b;
    for (i = (0); i < (10); ++i) {
      if (mp[i] == 1 and a == INF) {
        a = i;
      } else if (mp[i] == 0) {
        a = min(a, i);
        b = i;
        break;
      }
    }
    cout << a + b << "\n";
  }
  return 0;
}
bool is_prime(long long long int num) {
  return (num > 1 &&
          (num == 2 || ((((num)&1LL) ? (1) : (0)) &&
                        !(marked[(num) >> 6] & (1L << (((num) % 64) >> 1))))));
}
void sieve(long long long int n) {
  long long long int i, j;
  (marked[(1) >> 6] |= (1L << (((1) % 64) >> 1)));
  for (i = 3; i * i <= n; i += 2) {
    if (!(marked[(i) >> 6] & (1L << (((i) % 64) >> 1)))) {
      for (j = i * i; j <= n; j += i + i)
        (marked[(j) >> 6] |= (1L << (((j) % 64) >> 1)));
    }
  }
  prime.clear();
  prime.push_back(2);
  for (i = 3; i <= n; i += 2) {
    if (is_prime(i)) prime.push_back(i);
  }
}
