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 
#include <bits/stdc++.h>
int p[222], vis[222], x[111];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    memset(vis, 0, sizeof(vis));
    int n, i, e;
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
      scanf("%d", &e);
      x[i] = e;
      vis[e] = 1;
    }
    int j, f = 0;
    for (i = 1; i <= n; i++) {
      p[2 * i - 1] = x[i];
      f = 0;
      for (j = x[i] + 1; j <= 2 * n; j++) {
        if (vis[j] == 0) {
          f = 1;
          p[2 * i] = j;
          vis[j] = 1;
          break;
        }
      }
      if (f == 0) {
        break;
      }
    }
    if (f == 0)
      printf("-1\n");
    else {
      for (i = 1; i <= 2 * n; i++) {
        printf("%d", p[i]);
        if (i == 2 * n)
          printf("\n");
        else
          printf(" ");
      }
    }
  }
  return 0;
}
