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 JAVA):

Solution:
import java.util.*;
import java.io.*;

public class c {
  static boolean debug = false;
  public static void main(String[] args) {
    FS in = new FS(System.in);
    PrintWriter out = new PrintWriter(System.out);
    
    int numT = in.nextInt();
    for(int t=0; t<numT; t++) {
      int n = in.nextInt(); 

      int[] b = new int[n];
      int[] bvalLocs = new int[2*n+1];
      boolean[] used = new boolean[2*n+1];
      Arrays.fill(bvalLocs, -1);
      for(int i=0; i<n; i++) {
        b[i] = in.nextInt(); 
        bvalLocs[b[i]] = i;
        used[b[i]] = true;
      }

      if(debug) System.out.println(Arrays.toString(used));

      // Impossible cases = 
      // a[i] = max(b[2*i-1], b[2])
      // If not enough numbers (used up already)
      // If 1 isn't there lol.

      // for(int bind=n-1; bind>=0; bind--) {
      //   int aind = bind*2;

      // }

      int[] a = new int[n*2];
      int aind = 0;
      boolean impossible = false;
      // for(int bval=1; bval<=2*n; bval++) {
      for(int bind=0; bind<n; bind++) {
        // Search for it
        // int bind = bvalLocs[bval];
        // if(bind == -1) continue;
        int bval = b[bind];
        if(debug) System.out.printf("bval=%d bind=%d\n", bval, bind);
        
        // So now we have the correct thing.
        // Greedily place the next highest number.
        aind = bind*2;
        a[aind] = bval; // This is the min of the two, will always be first
        aind++;
        int nextUse = -1;
        for(int i=bval; i<=2*n; i++) {
          if(!used[i]) {
            // This is the next value we place.
            nextUse = i;
            break;
          }
        }
        
        if(debug) System.out.printf("Placing [%d]=%d\n", aind, nextUse);

        // If we can't place a next one, impossible
        if(nextUse == -1) {
          impossible = true;
          break;
        }

        a[aind] = nextUse; 
        aind++;
        used[nextUse] = true;
        

      }

      if(impossible) {
        out.println("-1");
      } else {
        for(int i=0; i<2*n; i++) {
          out.print(a[i] + " ");
        }
        out.println();
      }
      out.flush();

    }
  }
  static class FS {
    BufferedReader in;
    StringTokenizer token;
    public FS(InputStream stream) {
      in = new BufferedReader(new InputStreamReader(stream));
    }
    public String next() {
      if(token == null || !token.hasMoreElements()) {
        try{
          token = new StringTokenizer(in.readLine());
        } catch (Exception e) {}
        return next();
      }
      return token.nextToken();
    }
    public int nextInt() {
      return Integer.parseInt(next());
    }
  }
}

/*

5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8


*/