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;
const long long N = 2e2 + 5, mod = 1e9 + 7, mod1 = 998244353, mod2 = 1e9 + 9,
                inf = 1e18 + 7;
const long long infll = 1e18 + 7;
long long n, m, k;
long long ans, cnt, tmp, idx, mx = -inf, mn = inf;
long long x, y, z;
string s, t;
long long a[N], b[N], c[N];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long q = 1;
  cin >> q;
  while (q--) {
    bool ck = 0;
    memset(c, 0, sizeof(c));
    memset(a, 0, sizeof(a));
    cin >> n;
    set<long long> stt;
    for (long long i = 1; i <= 2 * n; i++) {
      c[i] = 1;
    }
    for (long long i = 1; i <= n; i++) {
      cin >> b[i];
      if (!c[b[i]]) {
        ck = 1;
        break;
      }
      c[b[i]] = 0;
      a[2 * i - 1] = b[i];
    }
    if (ck) {
      cout << -1 << '\n';
      continue;
    }
    for (long long i = 1; i <= 2 * n; i++) {
      if (c[i]) {
        bool ckk = 1;
        for (long long j = 1; j <= n; j++) {
          if (b[j] < i && !a[j * 2]) {
            ckk = 0;
            a[j * 2] = i;
            break;
          }
        }
        if (ckk) {
          ck = 1;
          break;
        }
      }
    }
    if (ck) {
      cout << -1 << '\n';
      continue;
    }
    for (long long iiii = 1; iiii <= 2 * n; iiii++) cout << a[iiii] << ' ';
    cout << '\n';
    ;
  }
}