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>
#pragma GCC optimize("Ofast")
using namespace std;
const long double PI = acos(-1.0L);
const long long MOD = 1000000007LL;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
void solve() {
  int N;
  cin >> N;
  vector<int> B(N);
  for (int i = 0; i < (int)(N); ++i) cin >> B[i];
  set<int> ppp, dic;
  for (int i = 0; i < (int)(N); ++i) ppp.emplace(B[i]);
  for (int i = 0; i < (int)(2 * N); ++i)
    if (!ppp.count(i + 1)) dic.emplace(i + 1);
  vector<int> ans(2 * N);
  for (int i = 0; i < (int)(N); ++i) {
    ans[i * 2] = B[i];
    bool ok = false;
    for (auto j : dic) {
      if (j > B[i]) {
        ans[i * 2 + 1] = j;
        ok = true;
        break;
      }
    }
    if (!ok) {
      cout << -1 << '\n';
      return;
    } else
      dic.erase(ans[i * 2 + 1]);
  }
  for (int i = 0; i < (int)(2 * N); ++i) cout << ans[i] << " ";
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int Q;
  cin >> Q;
  while (Q--) {
    solve();
  }
}