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;
 
#define online_judge            freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
#define fastio                    ios::sync_with_stdio(false);cin.tie(0);
#define ll                      long long
#define pii                     pair<int, int>
#define pli                     pair<ll, int>
#define pil                     pair<int, ll>
#define pll                     pair<ll, ll>
#define vi                      vector<int>
#define vll                     vector<ll>
#define vb                      vector<bool>
#define vd                      vector<double>
#define vs                      vector<string>
#define ff                      first
#define ss                      second
#define pb                      push_back
#define eb                      emplace_back
#define ppb                     pop_back
#define pf                      push_front
#define ppf                     pop_front
#define vpii                    vector<pii>
#define vpll                    vector<pll>
#define umap                    unordered_map
#define all(x)                  x.begin(),x.end()
#define clr(a,b)                memset(a,b,sizeof a)
#define fr(i, n)                for(ll i=0; i<n;++i)
#define fr1(i, n)               for(ll i=1; i<=n; ++i)
#define rfr(i, n)               for(ll i=n-1; i>=0; --i)
#define precise(x)              cout<<fixed<<setprecision(x)
#define wh                      while
#define mp                      make_pair
#define le                      length()
#define endl                    "\n"
#define sf                        scanf    
 
const ll M=1e9+7; 
 
ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
 
ll lcm(ll a,ll b)
{ return (a*b)/__gcd(a,b); }
 
 
ll power(ll x, ll y, ll p)  
{  
    ll res = 1;    
 
    x = x % p;
   
    if (x == 0) return 0;
 
    while (y > 0)  
    {  
       
        if (y & 1)  
            res = (res*x) % p;  
 
       
        y = y>>1; // y = y/2  
        x = (x*x) % p;  
    }  
    return res;  
}  
bool inside(ll x,ll y,ll n,ll m)
{
    return(x>=0&&x<n&&y>=0&&y<m);
}  

ll powmod(ll n)
{
    ll ret = 1;
    ll a = 2;
    while (n > 0) {
        if (n & 1) ret = ret * a % M;
        a = a * a % M;
        n >>= 1;
    }
    return ret;
}
 
ll modfact(ll n, ll p)
{
    if (n >= p)
        return 0;
 
    ll result = 1;
    for (int i = 1; i <= n; i++)
        result = (result * i) % p;
 
    return result;
}

long long binpower(long long a, long long b) {
    if (b == 0)
        return 1;
    long long res = binpower(a, b / 2);
    if (b % 2)
        return res * res * a;
    else
        return res * res;
}

///////////// TEMPLATE END /////////////////////////////////



void solve()
{
	int n;
	cin>>n;
	vll a(n);
	map<ll,ll> m;
	fr(i,n)
	{
		cin>>a[i];
		m[a[i]]++;
	}
	
	sort(all(a));
	ll mex1=0,mex2=0;
	ll i=0;
	bool c=true;
	fr(j,n+1)
	{
		if(m[j]>=2)
		{
			mex1++;
			mex2++;
		}
		
		else if(m[j]==0)
		{
			break;
		}
		else if(m[j]==1)
		{
			mex1++;
		}
	
	}
	cout<<mex1+mex2<<endl;
}



int main()
{
    fastio

    int t;
    cin>>t;
    fr1(i,t)
    {
//        cout<<"Case #"<<i<<": ";
        solve();
    }
   
    return 0;
}