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 int N = 205;
int b[N], p[N], a[N];
int n;
void check() {
  for (int i = 0; i < 2 * n; i += 2) {
    a[i] = b[i / 2];
    bool flag = false;
    for (int j = b[i / 2] + 1; j <= 2 * n; ++j) {
      if (p[j] == 0) {
        p[j] = 1, a[i + 1] = j;
        flag = true;
        break;
      }
    }
    if (!flag) {
      printf("-1\n");
      return;
    }
  }
  for (int i = 0; i < 2 * n; ++i) printf("%d ", a[i]);
  printf("\n");
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    scanf("%d", &n);
    memset(p, 0, sizeof(p));
    for (int i = 0; i < n; ++i) scanf("%d", &b[i]), p[b[i]] = 1;
    check();
  }
  return 0;
}