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 int N = 1e3 + 9;
const int M = 1e5 + 9;
inline int read() {
  int op = 1, x = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') op = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
  return x * op;
}
unordered_map<int, int> mp;
int vis[N], b[N];
struct node {
  int x, pos;
  friend bool operator<(node A, node B) { return A.x > B.x; }
} a[N];
int main() {
  int T, n;
  while (cin >> T) {
    while (T--) {
      memset(vis, 0, sizeof(vis));
      cin >> n;
      int mark = 0;
      for (int i = 1; i <= n; ++i) {
        cin >> a[i].x;
        if (a[i].x == 2 * n) mark = 1;
        vis[a[i].x] = 1;
        a[i].pos = i;
        b[i * 2 - 1] = a[i].x;
      }
      if (mark) {
        puts("-1");
        continue;
      }
      for (int i = 1; i <= n; ++i) {
        int ff = 0;
        for (int j = a[i].x + 1; j <= 2 * n; ++j)
          if (!vis[j]) {
            ff = 1;
            vis[j] = 1;
            b[i * 2] = j;
            break;
          }
        if (ff == 0) {
          mark = 1;
          break;
        }
      }
      if (mark) {
        puts("-1");
        continue;
      }
      for (int i = 1; i < 2 * n; ++i) cout << b[i] << " ";
      cout << b[2 * n] << '\n';
    }
  }
  return 0;
}