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;
int n, k;
bool vis[510];
struct node {
  int x, y;
} a[110];
int main() {
  cin >> k;
  while (k--) {
    memset(vis, 0, sizeof(vis));
    cin >> n;
    int ma = 2 * n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i].x;
      vis[a[i].x] = true;
    }
    int now, t = 1;
    for (int i = 1; i <= n; i++) {
      now = a[i].x;
      while (vis[now] == true || now <= a[i].x) {
        ++now;
      }
      a[i].y = now;
      if (now > ma) t = 0;
      vis[now] = true;
    }
    if (t == 0) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) cout << a[i].x << " " << a[i].y << " ";
    cout << endl;
  }
  return 0;
}