Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long N = 200 + 10;
const long long M = 1e5 + 10;
const long long inf = 1e9 + 7;
const long long Mod = 1e9 + 7;
const double eps = 1e-6;
int T;
int n;
int a[N], b[N];
bool fix[N];
bool f;
void init() {
  for (int i = 0; i < N; ++i) {
    fix[i] = 0;
  }
  f = 1;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    init();
    for (int i = 1; i <= n; ++i) {
      scanf("%d", b + i);
      a[2 * i - 1] = b[i];
      fix[b[i]] = 1;
    }
    for (int i = 2; i <= 2 * n; i += 2) {
      bool find = 0;
      for (int j = b[i / 2] + 1; j <= 2 * n; ++j) {
        if (fix[j]) continue;
        a[i] = j;
        fix[j] = 1;
        find = 1;
        break;
      }
      if (find == 0) {
        f = 0;
        break;
      }
    }
    if (f == 0)
      puts("-1");
    else {
      for (int i = 1; i <= 2 * n; ++i) printf("%d ", a[i]);
      puts("");
    }
  }
  return 0;
}