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

/*بِسْمِ اللَّهِ الرَّحْمَنِ الرَّحِيم*/

//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
typedef long long ll;
using pii = pair<int, int>;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;
//const ll mod = 998244353;


inline void normal(ll &a) { a %= mod; (a < 0) && (a += mod); }
inline ll modMul(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a * b) % mod; }
inline ll modAdd(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a + b) % mod; }
inline ll modSub(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while (p) { if (p & 1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, mod - 2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }

#define si(x) scanf("%d",&x)
#define sii(x,y) scanf("%d %d",&x,&y)
#define siii(x,y,z) scanf("%d %d %d",&x,&y,&z)
#define sl(x) scanf("%lld",&x)
#define sll(x,y) scanf("%lld %lld",&x,&y)
#define slll(x,y,z) scanf("%lld %lld %lld",&x,&y,&z)
#define ss(ch) scanf("%s",ch)
#define pi(x) printf("%d",x)
#define pii(x,y) printf("%d %d",x,y)
#define piii(x,y,z) printf("%d %d %d",x,y,z)
#define pl(x) printf("%lld",x)
#define pll(x,y) printf("%lld %lld",x,y)
#define plll(x,y,z) printf("%lld %lld %lld",x,y,z)
#define ps(ch) printf("%s",ch)
#define F(i,a,b)      for(int i= a; i <= b; i++)
#define R(i,b,a)      for(int i= b; i >= a; i--)
#define REP(i,n) for(int i = 0; i < (n); i++)

int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
int kx8[] = {1, 1, 2, 2, -1, -1, -2, -2};
int ky8[] = {2, -2, 1, -1, 2, -2, 1, -1};
/* for Random Number generate
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
*/
///**
template < typename F, typename S >ostream& operator << ( ostream& os, const pair< F, S > & p ) {return os << "(" << p.first << ", " << p.second << ")";}
template < typename T >ostream &operator << ( ostream & os, const vector< T > &v ) {os << "{"; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin() ) os << ", "; os << *it;} return os << "}";}
template < typename T >ostream &operator << ( ostream & os, const set< T > &v ) {os << "["; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin()) os << ", "; os << *it;} return os << "]";}
template < typename F, typename S >ostream &operator << ( ostream & os, const map< F, S > &v ) {os << "["; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin() ) os << ", "; os << it -> first << " = " << it -> second ;} return os << "]";}
#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)
clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)
void faltu () { cerr << endl; }
template <typename T>void faltu( T a[], int n ) {for (int i = 0; i < n; ++i) cerr << a[i] << ' '; cerr << endl;}
template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) { cerr << arg << ' '; faltu(rest...); }

// Program showing a policy-based data structure.
#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp>
#include <functional> // for less 
using namespace __gnu_pbds;

// GNU link : https://goo.gl/WVDL6g
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
        tree_order_statistics_node_update>
        new_data_set;
// find_by_order(k) – ফাংশনটি kth ordered element এর একটা পয়েন্টার রিটার্ন করে। অর্থাৎ তুমি চাইলেই kth ইন্ডেক্সে কি আছে, সেটা জেনে ফেলতে পারছো!
// order_of_key(x) – ফাংশনটি x এলিমেন্টটা কোন পজিশনে আছে সেটা বলে দেয়।
//*//**___________________________________________________**/
const int N = 1000006;


int main()
{
	FASTIO
	///*
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	freopen("error.txt", "w", stderr);
#endif
//*/
	int T;
	T = 1;
	//scanf("%d", &T);
	cin >> T;
	for (int cs = 1; cs <= T; cs++) {
		int n;
		cin >> n;
		vector<int> cnt(105);
		for (int i = 0; i < n; i++) {
			int x;
			cin >> x;
			cnt[x]++;
		}
		int a = 0, b = 0;
		bool flag = false;
		for (int i = 0; i <= 101; i++) {
			if (cnt[i] == 0) {
				break;
			}
			else if (cnt[i] == 2 && flag == false) {
				a++;
				b++;
			}
			else {
				flag = true;
				a++;
			}
		}
		//ans = max(ans, a + b);
		cout << a+b << "\n";
	}
	return 0;
}