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;
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const long double pi = 3.14159265359;
const long long mod = 1000000007;
const long long modc = 998244353;
const long long inf = LLONG_MAX;
const long long neg_inf = LLONG_MIN;
long long inline power(long long a, long long b, long long p) {
  a %= p;
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return ans;
}
long long mod_inv(long long n, long long p) { return power(n, p - 2, p); }
bool inline isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
template <typename... T>
void W(T &&...args) {
  ((cout << fixed << setprecision(17) << args << " "), ...);
  cout << endl;
}
template <typename... T>
void R(T &&...args) {
  ((cin >> args), ...);
}
long long ston(string s) {
  long long a = 0;
  reverse(s.begin(), s.end());
  long long pow = 1;
  for (long long i = 0; i < s.size(); i++) {
    a += pow * (s[i] - '0');
    pow *= 10;
  }
  return a;
}
template <class T>
string to_binary(T n) {
  string binaryNum;
  long long i = 0;
  while (n > 0) {
    binaryNum.push_back((n % 2) + '0');
    n = n / 2;
    i++;
  }
  if ((long long)binaryNum.size() == 0) binaryNum.push_back('0');
  reverse(binaryNum.begin(), binaryNum.end());
  return binaryNum;
}
long long bton(string &s, long long base) {
  long long ans = 0;
  long long pow = 1;
  reverse(s.begin(), s.end());
  for (auto x : s) {
    if (x >= 'a' && x <= 'z')
      ans += (x - 'a' + 10) * pow;
    else
      ans += (x - '0') * pow;
    pow *= base;
  }
  return ans;
}
template <class T>
void wv(T &v) {
  for (auto x : v) cout << x << " ";
  cout << endl;
}
template <class T>
void rv(T &v) {
  for (long long i = 0; i < (long long)v.size(); i++) cin >> v[i];
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
const long long M = 2e5;
void solve() {
  long long i, j, sum = 0, prod = 1, maxl = neg_inf, minl = inf;
  long long n;
  cin >> n;
  vector<long long> a(n);
  rv(a);
  unordered_map<long long, long long, custom_hash> m;
  for (auto x : a) m[x]++;
  long long first = -1, second = -1;
  for (i = 0; i <= 101; i++) {
    if (first == -1) {
      if (m[i] == 0) first = i;
    }
    if (second == -1) {
      if (m[i] <= 1) second = i;
    }
  }
  cout << first + second << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  R(t);
  while (t--) {
    solve();
  }
}
