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,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long double PI = acos(-1);
const int N = 1e5 + 10;
const int mod = 1e9 + 7;
const int mod1 = 998244353;
const long long INF = 2e18;
template <class T>
inline void amin(T& x, const T& y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T& x, const T& y) {
  if (x < y) x = y;
}
int read() {
  long long s = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    s = s * 10 + c - '0';
    c = getchar();
  }
  return s * f;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tt;
  tt = read();
  while (tt--) {
    int n;
    n = read();
    bool visited[210];
    memset(visited, 0, sizeof(visited));
    int a[105];
    vector<int> v;
    vector<int> t;
    bool flag = false;
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      t.push_back(a[i]);
      visited[a[i]] = true;
    }
    sort(t.begin(), t.end());
    for (int i = 1; i <= 2 * n; i++) {
      if (!visited[i]) {
        v.push_back(i);
        visited[i] = true;
      }
    }
    sort(v.begin(), v.end());
    for (int i = 0; i < n; i++) {
      if (t[i] > v[i]) {
        flag = true;
      }
    }
    if (flag) {
      cout << "-1\n";
      continue;
    } else {
      for (int i = 1; i <= n; i++) {
        cout << a[i] << " ";
        for (int j = 0; j < n; j++) {
          if (v[j] > a[i] && visited[v[j]]) {
            cout << v[j] << " ";
            visited[v[j]] = false;
            break;
          }
        }
      }
    }
    cout << "\n";
  }
  return 0;
}