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.io.*;
import java.util.*;

public class Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int t=Integer.parseInt(bu.readLine());
        while(t-->0)
        {
            int n=Integer.parseInt(bu.readLine());
            String s[]=bu.readLine().split(" ");
            int i,a[]=new int[n];
            for(i=0;i<n;i++) a[i]=Integer.parseInt(s[i]);

            int b[]=new int[n];
            if(n%2==0)
            {
                for(i=0;i<n;i+=2)
                {
                    b[i]=-a[i+1];
                    b[i+1]=a[i];
                }
                for(i=0;i<n;i++) sb.append(b[i]+" ");
                sb.append("\n");
                continue;
            }

            //a*(b+c)=ba+ca we need -ve of ba+ca
            //note that there may be a b+c which is 0, so we need to choose properly
            int j,x=-1,y=-1;
            for(i=0;i<n && x==-1;i++)
            for(j=i+1;j<n && x==-1;j++)
            if(a[i]+a[j]!=0)
            {
                x=i; y=j;
                break;
            }

            boolean vis[]=new boolean[n];
            vis[x]=true;
            vis[y]=true;
            int z=-1;
            for(i=0;i<n;i++)
            if(!vis[i]) {z=i; break;}

            b[z]=a[x]+a[y];
            b[x]=-a[z];
            b[y]=-a[z];
            vis[z]=true;
            for(i=0;i<n;i++)
            if(!vis[i])
            {
                x=i; vis[i]=true;
                y=i+1;
                while(y<n && vis[y]) y++;
                vis[y]=true;
                b[x]=a[y];
                b[y]=-a[x];
            }

            for(i=0;i<n;i++) sb.append(b[i]+" ");
            sb.append("\n");
        }
        System.out.print(sb);
    }
}