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;
const long long int INF = 1LL << 60;
const long long int N = 1e5 + 5;
long long int n, m, k, x, y, z, q;
int mod = 1e9 + 7;
long long int add(long long int a, long long int b) {
  if (a + b >= mod) return a + b - mod;
  return a + b;
}
long long int sub(long long int a, long long int b) {
  if (a < b) return a - b + mod;
  return a - b;
}
long long int mul(long long int a, long long int b) {
  return 1LL * a * b % mod;
}
long long int power(long long int a, long long int b) {
  if (b < 0) return 0;
  int re = 1;
  while (b) {
    if (b & 1) {
      re = mul(re, a);
    }
    a = mul(a, a);
    b >>= 1;
  }
  return re;
}
bool sortbysec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
bool sortbyfi(
    const pair<long long int, pair<long long int, long long int>> &a,
    const pair<long long int, pair<long long int, long long int>> &b) {
  return (a.first > b.first);
}
priority_queue<pair<long long int, long long int>,
               vector<pair<long long int, long long int>>,
               greater<pair<long long int, long long int>>>
    ppq;
bool bs(long double xx) {
  long long int sb = xx / (x * y);
  long long int sb1 = xx / x;
  long long int sb2 = xx / y;
  if ((xx - sb1) < n or (xx - sb2) < m) return false;
  xx -= sb;
  long long int sum = m + n;
  return sum <= xx;
}
bool cmp1(std::pair<long long int, long long int> a,
          std::pair<long long int, long long int> b) {
  return a.second < b.second;
}
bool cmp2(std::pair<long long int, long long int> a,
          std::pair<long long int, long long int> b) {
  return a.second > b.second;
}
void Birdperson() {
  cin >> n;
  vector<int> aa(101, 0);
  for (int i = 0; i < (n); ++i) {
    cin >> x;
    aa[x]++;
  }
  int ans = 0;
  for (int i = 0; i < (101); ++i)
    if (aa[i] < 2) {
      ans += i;
      break;
    }
  for (int i = 0; i < (101); ++i)
    if (aa[i] < 1) {
      ans += i;
      break;
    }
  cout << ans << "\n";
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  long long int ct = 1;
  cin >> t;
  while (ct <= t) {
    Birdperson();
    ++ct;
  }
}
