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 = 105;
int T, n, tot;
int a[N], b[N << 1];
bool vis[N << 1];
int main() {
  scanf("%d", &T);
  while (T--) {
    memset(vis, false, sizeof(vis));
    scanf("%d", &n);
    for (register int i = 1; i <= n; ++i) scanf("%d", &a[i]), vis[a[i]] = true;
    bool jay = true;
    for (register int i = 1; i <= n; ++i)
      if (a[i] == 2 * n) {
        jay = false;
        break;
      }
    if (!jay) {
      puts("-1");
      continue;
    }
    int tot = 0;
    for (register int i = 1; i <= n; ++i) {
      bool flag = false;
      for (register int j = a[i] + 1; j <= 2 * n; ++j)
        if (!vis[j]) {
          b[++tot] = a[i];
          b[++tot] = j;
          vis[a[i]] = vis[j] = true;
          flag = true;
          break;
        }
      if (!flag) {
        jay = false;
        break;
      }
    }
    if (!jay) {
      puts("-1");
      continue;
    }
    for (register int i = 1; i <= tot; ++i) printf("%d ", b[i]);
    puts("");
  }
  return 0;
}