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.io.*;
  
  public class MyClass {
      
      public static void main(String args[]) throws Exception {
        BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
        
        int T = Integer.parseInt(s.readLine());
        
        for (int t = 0; t < T; ++t) {
            int n = Integer.parseInt(s.readLine());
            
            int[] a = new int[n + 5];
            int[] b = new int[n + 5];
            String[] ss = s.readLine().split(" ");
            
            for (int i = 0; i < n; ++i) {
                a[i] = Integer.parseInt(ss[i]);
            }
            
            if (n % 2 == 1) {
                int Q = solve(a[n - 3], a[n - 2] + a[n - 1]);
                for (int k = 1; k <= 3; ++k) {
                    int c = a[n - k];
                    int d = 0;
                    for (int p = 1; p <= 3; ++p) if (p != k) {
                        d += a[n - p];
                    }
                    
                    if (c == 0 || d == 0) {
                        continue;
                    }
                    
                    Q = solve(c, d);
                    for (int p = 1; p <= 3; ++p) if (p != k) {
                        b[n - p] = -Q / d;
                    } else {
                        b[n - p] = Q / c;
                    }
                }
                
                
                for (int i = 0; i < n - 3; i += 2) {
                    Q = solve(a[i], a[i + 1]);
                    b[i] = Q / a[i];
                    b[i + 1] = -Q / a[i + 1];
                }
            } else {
               for (int i = 0; i < n; i += 2) {
                    int Q = solve(a[i], a[i + 1]);
                    b[i] = Q / a[i];
                    b[i + 1] = -Q / a[i + 1];
                }  
            }
            
            for (int i = 0; i < n - 1; ++i) {
                System.out.print(b[i] + " ");
            }
            System.out.println(b[n - 1]);
        }
      }
      
      private static int solve(int a, int b) {
          return a * b / gcd(a, b);
      }
      
      private static int gcd(int a, int b) {
          return (b == 0) ? a : gcd(b, a % b);
      }
  }