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 Main {
  	
  	//----------- StringBuilder for faster output------------------------------
  	static StringBuilder out = new StringBuilder();
  
  
  	public static void main(String[] args) {
  		FastScanner fs=new FastScanner();
  
  		/******  CODE STARTS HERE  *****/
  		
  		int t = fs.nextInt();
  		while(t-->0) {
  			int n = fs.nextInt();
  			int[] a = fs.readArray(n);
  			int[] b = new int[n];
  			if(n%2==0) {
  				for(int i=0; i<n; i+=2) {
  					b[i] = -a[i+1];
  					b[i+1] = a[i];
  				}
  			}else {
  				for(int i=0; i<=n-4; i+=2) {
  					b[i] = -a[i+1];
  					b[i+1] = a[i];
  				}
  				int[][] pre = new int[][] {{a[n-3], n-3}, {a[n-2], n-2}, {a[n-1], n-1}};
  				Arrays.sort(pre, (o1, o2) -> Integer.compare(o1[0], o2[0]));
  //				for(int i=0; i<3; i++) {
  //					System.out.println(pre[i][0]+"  "+pre[i][1]);
  //				}
  				if(pre[0][0]<0 && pre[1][0]<0) {
  					b[pre[0][1]] = b[pre[1][1]] = pre[2][0];
  					b[pre[2][1]] = -(pre[1][0]+pre[0][0]);
  				}else {
  					b[pre[1][1]] = b[pre[2][1]] = pre[0][0];
  					b[pre[0][1]] = -(pre[1][0]+pre[2][0]);
  				}
   			}
  			for(int e: b)out.append(e+" ");
  			out.append("\n");
   		}
  		System.out.print(out);
  	}
  	
  	static boolean isLimit(int[] a, int[] cnt) {
  		for(int x: a)if(cnt[x] != x)return false;
  		
  		return true;
  	}
  	
  	static int[] count(int[] a) {
  		int[] cnt = new int[a.length+1];
  		for(int x: a)cnt[x]++;
  		return cnt;
  	}
      
  	static void sort(int[] a) {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  	
  	
  	//----------- FastScanner class for faster input---------------------------
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
  	
  }