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 long long read() {
  char c = getchar();
  long long f = 1, x = 0;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ '0');
    c = getchar();
  }
  return x * f;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int b[1008], cnt[1008], t;
vector<int> ans;
int main() {
  scanf("%d", &t);
  while (t--) {
    int n, flag = 0;
    scanf("%d", &n);
    for (int i = 1; i <= 2 * n; i++) cnt[i] = 0;
    ans.clear();
    for (int i = 1; i <= n; i++) {
      scanf("%d", &b[i]);
      cnt[b[i]]++;
    }
    for (int i = 1; i <= n; i++) {
      flag = 0;
      ans.push_back(b[i]);
      for (int j = b[i] + 1; j <= 2 * n; j++) {
        if (!cnt[j]) {
          cnt[j]++;
          ans.push_back(j);
          flag = 1;
          break;
        }
      }
      if (!flag) {
        printf("-1\n");
        break;
      }
    }
    if (!flag) continue;
    for (int j = 0; j < ans.size(); j++) printf("%d ", ans[j]);
    printf("\n");
  }
  return 0;
}