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 yo {
  	public static void main(String[] args) throws Exception {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		int t = sc.nextInt();
  		for (int xx = 0; xx < t; xx++) {
  			int n = sc.nextInt();
  			int[] arr = new int[n];
  			for (int i = 0; i < n; i++) {
  				arr[i] = sc.nextInt();
  			}
  			if (n % 2 == 0) {
  				for (int i = 0; i < n; i = i + 2) {
  					pw.append(-arr[i + 1] + " " + arr[i] + " ");
  				}
  			} else {
  				int a = 0;
  				int b = 1;
  				int c = 2;
  				if (arr[a] >= 0 && arr[b] >= 0 && arr[c] >= 0) {
  					pw.append(-(arr[b] + arr[c]) + " " + arr[a] + " " + arr[a]+" ");
  				} else if (arr[a] < 0 && arr[b] < 0 && arr[c] < 0) {
  					pw.append(Math.abs(arr[b] + arr[c]) + " " + arr[a] + " " + arr[a]+" ");
  				} else if ((arr[a] >= 0 && arr[b] < 0 && arr[c] < 0) || (arr[a] < 0 && arr[b] >= 0 && arr[c] < 0) || (arr[a] < 0 && arr[b] < 0 && arr[c] >= 0)) {
  					if (arr[a] >= 0)
  						pw.append((arr[b] + arr[c]) + " " + -arr[a] + " " + -arr[a]);
  					else if (arr[b] >= 0)
  						pw.append(-arr[b] + " " + (arr[a] + arr[c]) + " " + -arr[b]);
  					else
  						pw.append(-arr[c] + " " + -arr[c] + " " + (arr[a] + arr[b]));
  					pw.append(" ");
  				} else {
  					if (arr[a] < 0)
  						pw.append(-(arr[b] + arr[c]) + " " + arr[a] + " " + arr[a]);
  					else if (arr[b] < 0)
  						pw.append(arr[b] + " " + -(arr[a] + arr[c]) + " " + arr[b]);
  					else
  						pw.append(arr[c] + " " + arr[c] + " " + -(arr[a] + arr[b]));
  					pw.append(" ");
  				}
  				for (int i = c + 1; i < n; i = i + 2) {
  					pw.append(-arr[i + 1] + " " + arr[i] + " ");
  				}
  			}
  			pw.append("\n");
  		}
  		pw.flush();
  	}
  }