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 t, n, m, k, ccnt, pos = 0, sum = 0, minn2 = INT_MAX, sum2, cnt2 = 0,
                            cnt3 = 0, cnt4 = 0, cnt5 = 0, cnt6 = 0, cnt7 = 0,
                            vis[100005], x1, y7, x2, y2, x3, y3, aaa, bbb;
long long a[1000005], b[1000005], c[1000005], c3[1000005], c4[1000005],
    kk[1000005], cnt = 0, x, y, z, d = 0, minn = LONG_LONG_MAX,
                 maxx = -LONG_LONG_MAX;
long long pos1 = 3, pos2 = 3, tem, tem2, l, r, ans = 0, ans2, mod = 1000000007,
          tw[100005], th[105], dd, ee, out[1000005], dp[5005][5005];
string s1, s2, s;
char cc[1005][1005], ch;
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  tw[0] = 1;
  for (int i = 1; i <= 55; i++) {
    tw[i] = tw[i - 1] * 2;
  }
  cin >> t;
  while (t--) {
    d = 1;
    cin >> n;
    for (int i = 0; i <= 2 * n; i++) {
      b[i] = 0;
    }
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      b[a[i]] = 1;
    }
    for (int i = 0; i < n; i++) {
      c[2 * i] = a[i];
      ans = -1;
      for (int j = a[i] + 1; j <= 2 * n; j++) {
        if (b[j] == 0) {
          ans = j;
          break;
        }
      }
      if (ans == -1) {
        d = 0;
        break;
      } else {
        c[2 * i + 1] = ans;
        b[ans] = 1;
      }
    }
    if (d) {
      for (int i = 0; i < 2 * n; i++) {
        cout << c[i] << " ";
      }
      cout << '\n';
    } else {
      cout << -1 << '\n';
    }
  }
}