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 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());
		}
	}
	
}