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 
import sys,os,io
# input = sys.stdin.readline # for strings
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline # for non-strings

PI = 3.141592653589793238460
INF =  float('inf')
MOD  = 1000000007
# MOD = 998244353


def bin32(num):
    return '{0:032b}'.format(num)

def add(x,y):
    return (x+y)%MOD

def sub(x,y):
    return (x-y+MOD)%MOD

def mul(x,y):
    return (x*y)%MOD

def gcd(x,y):
    if y == 0:
        return x
    return gcd(y,x%y)

def lcm(x,y):
    return (x*y)//gcd(x,y)

def power(x,y):
    res = 1
    x%=MOD
    while y!=0:
        if y&1 :
            res = mul(res,x)
        y>>=1
        x = mul(x,x)
        
    return res
        
def mod_inv(n):
    return power(n,MOD-2)

def prob(p,q):
    return mul(p,power(q,MOD-2))    
  
def ii():
    return int(input())

def li():
    return [int(i) for i in input().split()]

def ls():
    return [i for i in input().split()]

for t in range(ii()):
    n = ii()
    a = li()
    b = [0 for i in range(2*n)]

    for i in range(n):
        b[2*i] = a[i]
    # print(b)

    inds = [-1 for i in range(2 * n + 1)]
    for i in range(n):
        inds[a[i]] = i
    a.sort()
    a.reverse()
    lol = 1
    for i in a:
        ind = inds[i]
        flag = 0
        for j in range(2*n , i, -1):
            if inds[j ] == -1:
                flag = j
                
        if flag == 0:
            lol = 0
            break
        b[2*ind + 1] = flag
        inds[flag] = 2*ind + 1
    if lol:
        # print(b)
        for i in range(n):
            for j in range(i + 1, n):
                if b[2*i + 1] > b[2*j] and b[2*j + 1] > b[2*i] and b[2*i + 1] > b[2*j+1]:
                    b[2*i + 1] ,b[2*j + 1] = b[2*j + 1] , b[2*i + 1]
                    # print(i , j)

        print(*b)
    else:
        print(-1)
