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.util.*;
  import java.io.*;
  public class Main{
      static Scanner sc=new Scanner(System.in);
      static PrintWriter pw=new PrintWriter(System.out);
      public static void sort(int[]in) {
  		shuffle(in);
  		Arrays.sort(in);
  	}
  	public static void shuffle(int[]in) {
  		for(int i=0;i<in.length;i++) {
  			int idx=(int)(Math.random()*in.length);
  			int tmp=in[i];
  			in[i]=in[idx];
  			in[idx]=tmp;
  		}
  	}
      public static void main(String[] args) throws IOException, InterruptedException {
         int t=sc.nextInt();
         while(t-->0) {
      	   int n=sc.nextInt();
      	   int []arr=sc.nextIntArray(n);
      	   sort(arr);
      	   for(int i=0;i<n/2;i++) {
      		   pw.println(arr[n-1-i]+" "+arr[0]);
      	   }
         }
      	
      	
      	
      	
      	
          pw.flush();
      }
   
      static class pair implements Comparable<pair> {
          long x;
          long y;
   
          public pair(long x, long y) {
              this.x = x;
              this.y = y;
          }
   
          public String toString() {
              return x + " " + y;
          }
   
          public boolean equals(Object o) {
              if (o instanceof pair) {
                  pair p = (pair) o;
                  return p.x == x && p.y == y;
              }
              return false;
          }
   
          public int hashCode() {
              return new Double(x).hashCode() * 31 + new Double(y).hashCode();
          }
   
          public int compareTo(pair other) {
              if (this.x == other.x) {
                  return Long.compare(other.y, this.y);
              }
              return Long.compare(this.x, other.x);
          }
      }
      static class Scanner {
          StringTokenizer st;
          BufferedReader br;
          
   
          public Scanner(InputStream s) {
              br = new BufferedReader(new InputStreamReader(s));
          }
   
          public Scanner(FileReader r) {
              br = new BufferedReader(r);
          }
          public String readAllLines(BufferedReader reader) throws IOException {
              StringBuilder content = new StringBuilder();
              String line;
   
              while ((line = reader.readLine()) != null) {
                  content.append(line);
                  content.append(System.lineSeparator());
              }
   
              return content.toString();
          }
          public String next() throws IOException {
              while (st == null || !st.hasMoreTokens())
                  st = new StringTokenizer(br.readLine());
              return st.nextToken();
          }
   
          public int nextInt() throws IOException {
              return Integer.parseInt(next());
          }
   
          public long nextLong() throws IOException {
              return Long.parseLong(next());
          }
   
          public String nextLine() throws IOException {
              return br.readLine();
          }
   
          public double nextDouble() throws IOException {
              String x = next();
              StringBuilder sb = new StringBuilder("0");
              double res = 0, f = 1;
              boolean dec = false, neg = false;
              int start = 0;
              if (x.charAt(0) == '-') {
                  neg = true;
                  start++;
              }
              for (int i = start; i < x.length(); i++)
                  if (x.charAt(i) == '.') {
                      res = Long.parseLong(sb.toString());
                      sb = new StringBuilder("0");
                      dec = true;
                  } else {
                      sb.append(x.charAt(i));
                      if (dec)
                          f *= 10;
                  }
              res += Long.parseLong(sb.toString()) / f;
              return res * (neg ? -1 : 1);
          }
   
          public long[] nextlongArray(int n) throws IOException {
              long[] a = new long[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextLong();
              return a;
          }
   
          public Long[] nextLongArray(int n) throws IOException {
              Long[] a = new Long[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextLong();
              return a;
          }
   
          public int[] nextIntArray(int n) throws IOException {
              int[] a = new int[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextInt();
              return a;
          }
   
          public Integer[] nextIntegerArray(int n) throws IOException {
              Integer[] a = new Integer[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextInt();
              return a;
          }
   
          public boolean ready() throws IOException {
              return br.ready();
          }
   
      }
   
  }
  