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.
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
 
 
import java.io.*;
 
public class Div3 {
	
	private static class MyScanner {
	    BufferedReader br;
	    StringTokenizer st;
 
	    public MyScanner() {
	       br = new BufferedReader(new InputStreamReader(System.in));
	    }
 
	    String next() {
	        while (st == null || !st.hasMoreElements()) {
	            try {
	                st = new StringTokenizer(br.readLine());
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	        }
	        return st.nextToken();
	    }
 
	    int nextInt() {
	        return Integer.parseInt(next());
	    }
 
	    long nextLong() {
	        return Long.parseLong(next());
	    }
 
	    double nextDouble() {
	        return Double.parseDouble(next());
	    }
 
	    String nextLine(){
	        String str = "";
		  try {
		     str = br.readLine();
		  } catch (IOException e) {
		     e.printStackTrace();
		  }
		  return str;
		  
	        }
	    }
		
	public static long gcd(long a, long b)
	{
		if(b==0)
			 return a;
		
		return gcd(b,a%b);
		
	}
	
	public static long lcm(long a, long b)
	{
		return (a*b)/gcd(a,b);
	}
	
	public static void solution (int n, long[] arr) {
		
		long[] ans = new long[n];
		for(int i = 0; i<n/2; i++)
			ans[i] = arr[n-1-i];
		for(int i = n/2; i<n; i++)
			ans[i] = -1*arr[n-1-i];
			
		if(n%2==1)
		{
			ans[n/2] = -1*(arr[n/2-1]+arr[n/2+1]);
			ans[n/2-1] = arr[n/2];
			ans[n/2+1] = arr[n/2];
		}
		
		for(int i = 0; i<n; i++)
			out.print(ans[i]+" ");
		
		out.println();
	}
	
					
private static PrintWriter out = new PrintWriter(System.out);
 
public static void main (String[] args)
  {
	MyScanner s =  new MyScanner();
	  
    int t = s.nextInt();
    
    while(t-->0)
    {
		int n = s.nextInt();
		long[] arr = new long[n];
		for(int i = 0; i<n; i++)
			arr[i] = s.nextLong();
		
        solution(n,arr);
    }
    
    out.flush();
    out.close();
    
}
}