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.*;
public class classs43 {
 
	public static void main(String[] args) {
		Scanner input=new Scanner(System.in);
		int t=input.nextInt();
		while(t-->0) {
			int n=input.nextInt();
	        long a[]=new long[n];
	        long sum=0;
	        for(int i=0;i<n;i++) {
	        	a[i]=input.nextLong();
	        	sum=sum+a[i];
	        }
	       sum=sum-a[n-1];
	       if(sum!=0) {
	    	   long s=sum*(-1);
	    	   for(int i=0;i<n;i++) {
	    		   if(i<n-1) {
	    		   System.out.print(a[n-1]+" ");
	    		   }
	    		   else {
	    			   System.out.print(s+" ");
	    		   }
	    	   }
	       }else {
	    	   sum=sum-a[n-2];
	    	   long s=a[n-1];
	    	   for(int i=0;i<n;i++) {
	    		   if(i<n-2) {
	    			   System.out.print(Math.abs(a[n-1])+" ");
	    		   }
	    		   else {
	    			   if(i==n-2) {
	    				   if(a[n-1]>0) {
	    					   s=s*(-1);
	    				   }
	    			   System.out.print(s+" ");
	    			   }
	    			   else {
	    				sum=sum*Math.abs(a[n-1]);
	    				sum=sum+(s*a[n-2]);
	    				s=sum/a[n-1];
	    				if(sum>0) {
	    				if(a[n-1]<0) {
	    					System.out.print(-s+" ");
	    				}
	    				else {
	    					System.out.print(s+" ");
	    				}
	    				} 
	    				else {
	    					if(a[n-1]>0) {
		    					System.out.print(s+" ");
		    				}
		    				else { 
		    					System.out.print(-s+" ");
		    				}
	    				}
	    			   }
	    		   }
	    	   }
	       }
			System.out.println();
		}
	}
 
}