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 <CodeforcesSolutions.h>
#include <ONLINE_JUDGE <solution.cf(contestID = "1406",questionID = "A",method = "GET")>.h>
"""
    Author : thekushalghosh
    Team   : CodeDiggers
 
    I prefer Python language over the C++ language :p :D
        
    Visit my website : thekushalghosh.github.io
"""
import sys,math,cmath,time
start_time = time.time()
##########################################################################
################# ---- THE ACTUAL CODE STARTS BELOW ---- #################
def solve():
    n = inp()
    a = inlt()
    q = counter(a)
    qw = 0
    for i in range(len(q)):
        if q[i] == 0:
            if qw == 0:
                qq = 2 * i
            else:
                qq = qq + i
            break
        if q[i] == 1 and qw == 0:
            if qw == 0:
                qq = i
                qw = 1
    print(qq)

################## ---- THE ACTUAL CODE ENDS ABOVE ---- ##################
##########################################################################
 
def main():
    global tt
    if not ONLINE_JUDGE:
        sys.stdin = open("input.txt","r")
        sys.stdout = open("output.txt","w")
    t = 1
    t = inp()
    for tt in range(t):
        solve()
    if not ONLINE_JUDGE:
        print("Time Elapsed :",time.time() - start_time,"seconds")
        sys.stdout.close()

#---------------------- USER DEFINED INPUT FUNCTIONS ----------------------#
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    return(input().strip())
def invr():
    return(map(int,input().split()))
 
#------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#
def counter(a):
    q = [0] * (max(a) + 2)
    for i in range(len(a)):
        q[a[i]] = q[a[i]] + 1
    return(q)

def counter_elements(a):
    q = dict()
    for i in range(len(a)):
        if a[i] not in q:
            q[a[i]] = 0
        q[a[i]] = q[a[i]] + 1
    return(q)
 
def string_counter(a):
    q = [0] * 26
    for i in range(len(a)):
        q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
    return(q)
 
def factors(n):
    q = []
    for i in range(1,int(n ** 0.5) + 1):
        if n % i == 0: q.append(i); q.append(n // i)
    return(list(sorted(list(set(q)))))
 
def prime_factors(n):
    q = []
    while n % 2 == 0: q.append(2); n = n // 2
    for i in range(3,int(n ** 0.5) + 1,2):
        while n % i == 0: q.append(i); n = n // i
    if n > 2: q.append(n)
    return(list(sorted(q)))
 
def transpose(a):
    n,m = len(a),len(a[0])
    b = [[0] * n for i in range(m)]
    for i in range(m): 
        for j in range(n): 
            b[i][j] = a[j][i]
    return(b)

def factorial(n,m = 1000000007):
    q = 1
    for i in range(n):
        q = (q * (i + 1)) % m
    return(q)
#-----------------------------------------------------------------------#
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
    input = sys.stdin.readline



main()