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;
long long INF = 1 << 28;
const double pi = acos(-1.0);
int fx[] = {1, -1, 0, 0};
int fy[] = {0, 0, 1, -1};
int dir[4][2] = {1, 0, -1, 0, 0, -1, 0, 1};
int knight[8][2] = {1, 2, 1, -2, 2, 1, 2, -1, -1, 2, -1, -2, -2, 1, -2, -1};
const long double EPS = 1e-7;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
bool cmp(int a, int b) { return a < b; }
int on(int mask, int st) { return mask | (1 << st); }
int off(int mask, int st) { return mask & (~(1 << st)); }
const int mx = 600100;
int mod = 998244353;
int vis[400];
pair<int, int> arr[mx];
int ans[1000];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    memset(ans, 0, sizeof(ans));
    memset(vis, 0, sizeof(vis));
    int n;
    int x;
    scanf("%d", &n);
    int f = 1;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &x);
      vis[x]++;
      ans[(i * 2) - 1] = x;
    }
    int l = 1;
    int r = 1;
    while (r <= 2 * n) {
      if (vis[r] == 0) {
        int flag = 0;
        for (int i = 1; i <= n; i++) {
          if (ans[(i * 2) - 1] < r && ans[i * 2] == 0) {
            flag = 1;
            ans[i * 2] = r;
            break;
          }
        }
        if (flag == 0) {
          f = 0;
          break;
        }
      }
      r++;
    }
    if (f == 0)
      cout << -1 << endl;
    else {
      for (int i = 1; i <= 2 * n; i++) printf("%d ", ans[i]);
      cout << endl;
    }
  }
  return 0;
}