description: Vupsen and Pupsen were gifted an integer array. Since Vupsen doesn't like the number 0, he threw away all numbers equal to 0 from the array. As a result, he got an array a of length n.
  
  Pupsen, on the contrary, likes the number 0 and he got upset when he saw the array without zeroes. To cheer Pupsen up, Vupsen decided to come up with another array b of length n such that ∑_{i=1}^{n}a_i ⋅ b_i=0. Since Vupsen doesn't like number 0, the array b must not contain numbers equal to 0. Also, the numbers in that array must not be huge, so the sum of their absolute values cannot exceed 10^9. Please help Vupsen to find any such array b!
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The next 2 ⋅ t lines contain the description of test cases. The description of each test case consists of two lines.
  
  The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5) — the length of the array.
  
  The second line contains n integers a_1, a_2, …, a_n (-10^4 ≤ a_i ≤ 10^4, a_i ≠ 0) — the elements of the array a.
  
  It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case print n integers b_1, b_2, …, b_n — elements of the array b (|b_1|+|b_2|+… +|b_n| ≤ 10^9, b_i ≠ 0, ∑_{i=1}^{n}a_i ⋅ b_i=0).
  
  It can be shown that the answer always exists.
  
  Example
  
  Input
  
  
  3
  2
  5 5
  5
  5 -2 10 -9 4
  7
  1 2 3 4 5 6 7
  
  
  Output
  
  
  1 -1
  -1 5 1 -1 -1
  -10 2 2 -3 5 -1 -1
  
  Note
  
  In the first test case, 5 ⋅ 1 + 5 ⋅ (-1)=5-5=0. You could also print 3 -3, for example, since 5 ⋅ 3 + 5 ⋅ (-3)=15-15=0
  
  In the second test case, 5 ⋅ (-1) + (-2) ⋅ 5 + 10 ⋅ 1 + (-9) ⋅ (-1) + 4 ⋅ (-1)=-5-10+10+9-4=0.
solution: import java.io.*;
  import java.util.*;
  
  public class Codeforces
  {
      public static void main(String args[])throws Exception
      {
          BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
          StringBuilder sb=new StringBuilder();
          int t=Integer.parseInt(bu.readLine());
          while(t-->0)
          {
              int n=Integer.parseInt(bu.readLine());
              String s[]=bu.readLine().split(" ");
              int i,a[]=new int[n];
              for(i=0;i<n;i++) a[i]=Integer.parseInt(s[i]);
  
              int b[]=new int[n];
              if(n%2==0)
              {
                  for(i=0;i<n;i+=2)
                  {
                      b[i]=-a[i+1];
                      b[i+1]=a[i];
                  }
                  for(i=0;i<n;i++) sb.append(b[i]+" ");
                  sb.append("\n");
                  continue;
              }
  
              //a*(b+c)=ba+ca we need -ve of ba+ca
              //note that there may be a b+c which is 0, so we need to choose properly
              int j,x=-1,y=-1;
              for(i=0;i<n && x==-1;i++)
              for(j=i+1;j<n && x==-1;j++)
              if(a[i]+a[j]!=0)
              {
                  x=i; y=j;
                  break;
              }
  
              boolean vis[]=new boolean[n];
              vis[x]=true;
              vis[y]=true;
              int z=-1;
              for(i=0;i<n;i++)
              if(!vis[i]) {z=i; break;}
  
              b[z]=a[x]+a[y];
              b[x]=-a[z];
              b[y]=-a[z];
              vis[z]=true;
              for(i=0;i<n;i++)
              if(!vis[i])
              {
                  x=i; vis[i]=true;
                  y=i+1;
                  while(y<n && vis[y]) y++;
                  vis[y]=true;
                  b[x]=a[y];
                  b[y]=-a[x];
              }
  
              for(i=0;i<n;i++) sb.append(b[i]+" ");
              sb.append("\n");
          }
          System.out.print(sb);
      }
  }