You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.

Input

Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).

The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).

The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).

It is guaranteed that the sum of n by all test cases doesn't exceed 100.

Output

For each test case, if there is no appropriate permutation, print one number -1.

Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.

Example

Input


5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8


Output


1 2 
-1
4 5 1 2 3 6 
-1
1 3 5 6 7 9 2 4 8 10 
class GFG: 
    def __init__(self,graph, seq): 
          
        # residual graph 
        self.graph = graph  
        self.ppl = len(graph) 
        self.jobs = len(graph[0])
        self.seq = seq
  
    # A DFS based recursive function 
    # that returns true if a matching  
    # for vertex u is possible 
    def bpm(self, u, matchR, seen): 
  
        # Try every job one by one 
        for v in range(self.jobs-1,-1,-1): 
  
            # If applicant u is interested  
            # in job v and v is not seen 
            if self.graph[u][v] and seen[v] == False: 
                  
                # Mark v as visited 
                seen[v] = True 
  
                '''If job 'v' is not assigned to 
                   an applicant OR previously assigned  
                   applicant for job v (which is matchR[v])  
                   has an alternate job available.  
                   Since v is marked as visited in the  
                   above line, matchR[v]  in the following 
                   recursive call will not get job 'v' again'''
                if matchR[v] == -1 or self.bpm(matchR[v],  
                                               matchR, seen): 
                    matchR[v] = u 
                    return True
        return False
  
    # Returns maximum number of matching  
    def maxBPM(self): 
        '''An array to keep track of the  
           applicants assigned to jobs.  
           The value of matchR[i] is the  
           applicant number assigned to job i,  
           the value -1 indicates nobody is assigned.'''
        matchR = [-1] * self.jobs 
          
        # Count of jobs assigned to applicants 
        result = 0 
        for i in self.seq: 
            i -= 1
            # Mark all jobs as not seen for next applicant. 
            seen = [False] * self.jobs 
              
            # Find if the applicant 'u' can get a job 
            if self.bpm(i, matchR, seen): 
                result += 1
        # print(matchR)
        self.matchR = matchR
        return result 
t = int(input())
for _ in range(t):
    n = int(input())
    b = list(map(int, input().split()))
    g = [[0 for _ in range(2*n)] for _ in range(2*n)]
    for i in b:
        for j in range(i,2*n):
            if j+1 not in b:
                g[i-1][j] = 1
    m = GFG(g, b)
    if m.maxBPM() != n:
        print(-1)
    else:
        for i in b:
            print(i, m.matchR.index(i-1)+1,  end=' ')
        print()

