description: You are given a sequence a_1, a_2, ..., a_n consisting of n pairwise distinct positive integers.
  
  Find \left⌊ \frac n 2 \right⌋ different pairs of integers x and y such that: 
  
    * x ≠ y; 
    * x and y appear in a; 
    * x~mod~y doesn't appear in a. 
  
  
  
  Note that some x or y can belong to multiple pairs.
  
  ⌊ x ⌋ denotes the floor function — the largest integer less than or equal to x. x~mod~y denotes the remainder from dividing x by y.
  
  If there are multiple solutions, print any of them. It can be shown that at least one solution always exists.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.
  
  The first line of each testcase contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of the sequence.
  
  The second line of each testcase contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6).
  
  All numbers in the sequence are pairwise distinct. The sum of n over all testcases doesn't exceed 2 ⋅ 10^5.
  
  Output
  
  The answer for each testcase should contain \left⌊ \frac n 2 \right⌋ different pairs of integers x and y such that x ≠ y, x and y appear in a and x~mod~y doesn't appear in a. Print the pairs one after another.
  
  You can print the pairs in any order. However, the order of numbers in the pair should be exactly such that the first number is x and the second number is y. All pairs should be pairwise distinct.
  
  If there are multiple solutions, print any of them.
  
  Example
  
  Input
  
  
  4
  2
  1 4
  4
  2 8 3 4
  5
  3 8 5 9 7
  6
  2 7 5 3 4 8
  
  
  Output
  
  
  4 1
  8 2
  8 4
  9 5
  7 5
  8 7
  4 3
  5 2
  
  Note
  
  In the first testcase there are only two pairs: (1, 4) and (4, 1). \left⌊ \frac 2 2 \right⌋=1, so we have to find one pair. 1~mod~4=1, and 1 appears in a, so that pair is invalid. Thus, the only possible answer is a pair (4, 1).
  
  In the second testcase, we chose pairs 8~mod~2=0 and 8~mod~4=0. 0 doesn't appear in a, so that answer is valid. There are multiple possible answers for that testcase.
  
  In the third testcase, the chosen pairs are 9~mod~5=4 and 7~mod~5=2. Neither 4, nor 2, appears in a, so that answer is valid.
solution: import java.io.*;
  import java.util.*;
  
  /**
   * -----------------|___________|---------------------
   * CCCCCCCCC   OOOOOOOOOO   DDDDDDDD      EEEEEEEEE
   * CCCC        OOO    OOO   DD    DDD     EEEE
   * CCCC        OOO    OOO   DD     DDD    EEEEEEEEEE
   * CCCC        OOO    OOO   DD     DDDD   EEEE
   * CCCCCCCCC   OOOOOOOOOO   DDDDDDDDDDDD  EEEEEEEEEE
   * -----------------|___________|---------------------
   */
  
  public class Main {
      static long M = 1000000007;
  
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          InputReader in = new InputReader(inputStream);
          PrintWriter sout = new PrintWriter(outputStream);
  
          int t = in.nextInt();
          while (t-- > 0) {
              int n = in.nextInt();
              int[] arr = new int[n];
              for(int i = 0; i<n; i++){
                  arr[i] = in.nextInt();
              }
              sortInt(arr);
              int cnt = 0;
              //List<Pair> pairs = new ArrayList<>();
              for(int i = 1; i<n && cnt<n/2; i++){
                  if(arr[i]!=arr[0]){
                      cnt++;
                      sout.println(arr[i]+" "+arr[0]);
                  }
              }
          }
          sout.close();
      }
  
      private static class Pair{
          int x;
          int y;
  
          public Pair(int x, int y) {
              this.x = x;
              this.y = y;
          }
      }
  
      public static long factorial(long N) {
          long result = 1;
          for (long x = 2; x <= N; x++) {
              result = (result * x) % M;
          }
          return result;
      }
  
      public static long binomial(long n) {
          long nom = factorial(n); // n!
          long denom = ((long) 2 * factorial(n - 2)) % M; // k!(n-k)!
          return (nom * fastPow(denom, M - 2)) % M;
      }
  
      public static long fastPow(long num, long pow) {
          if (pow == 0)
              return 1;
          if (pow % 2 == 0) {
              long result = fastPow(num, pow / 2);
              result = (result * result) % M;
              return result;
          }
          long result = fastPow(num, pow - 1);
          result = (result * num) % M;
          return result;
      }
  
  
      public static void sortLong(long[] a) {
          ArrayList<Long> l = new ArrayList<>();
          for (long i : a)
              l.add(i);
          Collections.sort(l);
          for (int i = 0; i < a.length; i++)
              a[i] = l.get(i);
      }
  
      public static void sortInt(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i : a)
              l.add(i);
          Collections.sort(l);
          for (int i = 0; i < a.length; i++)
              a[i] = l.get(i);
      }
  
      public static class Node {
          int value;
          boolean visit = false;
          List<Node> neighbors = new ArrayList<>();
  
          public Node(int value) {
              this.value = value;
          }
      }
  
      public static class InputReader {
          private final BufferedReader reader;
          private StringTokenizer tokenizer;
  
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 32768);
              tokenizer = null;
          }
  
          public String nextToken() {
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(nextToken());
          }
  
          public long nextLong() {
              return Long.parseLong(nextToken());
          }
  
          public double nextDouble() {
              return Double.parseDouble(nextToken());
          }
      }
  }