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;
inline int read() {
  int ans = 0, sign = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ans = ans * 10 + (ch - '0');
    ch = getchar();
  }
  return ans * sign;
}
const double PI = acos(-1.0);
const int mod = 1e9 + 7;
const int N = 100 + 5;
int a[N], b[N + N];
int vis[N + N];
void solve() {
  int n = read();
  bool flag = true;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
  }
  memset(b, 0, sizeof(b));
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    b[i * 2 - 1] = a[i];
    vis[a[i]] = 1;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = a[i] + 1; j <= 2 * n; j++)
      if (!vis[j] && j > a[i]) {
        b[i * 2] = j;
        vis[j] = 1;
        break;
      }
    if (!b[i * 2]) {
      flag = false;
      break;
    }
  }
  if (!flag)
    cout << "-1" << endl;
  else {
    for (int i = 1; i <= 2 * n - 1; i++) {
      cout << b[i] << " ";
    }
    cout << b[2 * n] << endl;
  }
}
int main() {
  int t = read();
  while (t--) {
    solve();
  }
  return 0;
}