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;
struct point {
  int x, y, ind;
  char ch;
};
long long mod = 1e9 + 7;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long fact[13];
void factorial() {
  fact[0] = 1;
  for (long long i = 1; i < 12; i++) fact[i] = (fact[i - 1] * i);
}
bool cmp(point A, point B) {
  if (A.x != B.x) return A.x < B.x;
  return A.y < B.y;
}
vector<long long> sie;
long long spf[1000007 + 7];
void sieve() {
  for (long long i = 2; i < 1000007; i++) {
    if (spf[i] == 0) {
      spf[i] = i;
      sie.push_back(i);
    }
    for (long long j = 0; j < ((long long)(sie).size()) &&
                          i * sie[j] <= 1000007 && sie[j] <= spf[i];
         j++)
      spf[i * sie[j]] = sie[j];
  }
}
const long long mx = 2e5 + 9;
long long n, a, b, _, ans, c, _1, _2, m, an, k, need, shuru, _3, _c, _a, _b,
    way, _0, mex, q, d, temp;
double DF, AF;
string st, good, orginal[2];
long long ar[mx], co[mx];
long long sol(long long x, long long y) { return _1; }
void f() {
  for (long long i = 0; i < 103; i++) co[i] = 0;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> _;
    co[_]++;
  }
  ans = 0;
  _ = 0;
  for (long long i = 0; i < 102; i++) {
    if (co[i] >= 2) continue;
    if (co[i] && _ == 0) {
      _ = 1;
      ans = i;
    } else if (co[i] == 0 && _ == 1) {
      _ = 2;
      ans += i;
    } else if (co[i] == 0 && _ == 0) {
      _ = 2;
      ans = i + i;
    }
    if (_ == 2) break;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(12);
  long long _t;
  cin >> _t;
  for (long long w = 0; w < _t; w++) f();
  return 0;
}
