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.util.*;
  import java.lang.*;
  import java.io.*;
  
  public class Main
  {
  	PrintWriter out = new PrintWriter(System.out);
  	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer tok = new StringTokenizer("");
      String next() throws IOException {
          if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
          return tok.nextToken();
      }
      int ni() throws IOException { return Integer.parseInt(next()); }
      long nl() throws IOException { return Long.parseLong(next()); }
      
      long mod=1000000007;
      
      void solve() throws IOException {
          for (int tc=ni();tc>0;tc--) {
              int n=ni();
              int[]A=new int[n];
              for (int i=0;i<n;i++) A[i]=ni();
              
              long[]B=new long[n];
              Arrays.fill(B,1);
              int s=0;
              int lowodd=1000000;
              int lop=-1;
              for (int i=0;i<n;i++) {
                  if (A[i]<0) { A[i]*=-1; B[i]=-1; }
                  s+=A[i];
                  if (A[i]%2==1 && A[i]<lowodd) { lowodd=A[i]; lop=i; }
              }
              
              if (s%2==1) { B[lop]*=2; s+=A[lop]; A[lop]*=2; }
              
              long lcm=Long.MAX_VALUE;
              int p=-1;
              
              for (int i=0;i<n;i++) {
                  long nl=s-A[i];
                  nl=(nl*A[i])/gcd(nl,A[i]);
                  if (nl<lcm) { lcm=nl; p=i; }
              }
              
              long mult=lcm/(s-A[p]);
              for (int i=0;i<n;i++) {
                  if (i==p) { B[i]*=lcm/A[i]; B[i]*=-1; }
                  else B[i]*=mult;
              }
              
              for (int i=0;i<n;i++) out.print(B[i]+" ");
              
              out.println();
          }
          out.flush();
      }
      
      int gcd(int a,int b) { return(b==0?a:gcd(b,a%b)); }
      long gcd(long a,long b) { return(b==0?a:gcd(b,a%b)); }
      long mp(long a,long p) { long r=1; while(p>0) { if ((p&1)==1) r=(r*a)%mod; p>>=1; a=(a*a)%mod; } return r; }
      
      public static void main(String[] args) throws IOException {
          new Main().solve();
      }
  }