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.Scanner;
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.BufferedWriter;
import java.io.OutputStreamWriter;

public class D {
	public static void main(String[] args) {
		try {
			BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
			BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));

			int t = Integer.parseInt(reader.readLine());
			for (; t > 0; t--) {
				int n = Integer.parseInt(reader.readLine());
				int[] a = new int[n];
				Integer[] c = new Integer[n];
				int[] b = new int[n];

				HashMap<Integer, LinkedList<Integer>> pos = new HashMap<>();

				String[] line = reader.readLine().split(" ");

				for (int i = 0; i < n; i++) {
					a[i] = Integer.parseInt(line[i]);
					c[i] = Integer.valueOf(a[i]);

					LinkedList<Integer> adjs = pos.getOrDefault(a[i], new LinkedList<>());
					adjs.add(i);
					pos.put(a[i], adjs);
				}

				Arrays.sort(c, (x, y) -> Math.abs(x) - Math.abs(y));

				int lim = n;
				if (n % 2 == 1)
					lim -= 3;

				for (int i = 0; i+1 < lim; i+=2) {
					Integer l = c[i]; Integer r = c[i+1];

					LinkedList<Integer> lAdjs = pos.get(l);
					LinkedList<Integer> rAdjs = pos.get(r);

					Integer lPos = lAdjs.get(0);
					lAdjs.remove(0);

					Integer rPos = rAdjs.get(0);
					rAdjs.remove(0);

					int g = Math.abs(gcd(r, l));
					b[lPos] = -r/g; b[rPos] = l/g;
				}

				int[] lp = new int[3];
				int lPos = 0;

				for (int i = 0; i < n; i++)
					if (b[i] == 0)
						lp[lPos++] = i;

				if (n % 2 == 1) {
					b[lp[0]] = a[lp[0]] * a[lp[2]];
					b[lp[1]] = a[lp[1]] * a[lp[2]];

					b[lp[2]] = -(a[lp[0]]*a[lp[0]] + a[lp[1]]*a[lp[1]]);
					int g = gcd(gcd(Math.abs(b[lp[0]]), Math.abs(b[lp[1]])), Math.abs(b[lp[2]]));
 
					for (int i = 0; i < 3; i++)
						b[lp[i]] /= g;
				}


				for (int i = 0; i < n; i++) {
					String s = String.valueOf(b[i]);
					writer.write(s, 0, s.length());
					writer.write(' ');
				}

				writer.flush();
			}

			writer.close();
			reader.close();
		} catch (Exception e) {
			e.printStackTrace();
		}
	}

	private static int gcd(int a, int b) {
		if (b == 0)	return a;
		return gcd(b, a%b);
	}

	private static void fill(int[] a, int[] b, int lim) {
		for (int i = 0; i < lim; i+=2) {
			int l = a[i]; int r = a[i+1];
			b[i] = -r; b[i+1] = l;

			int g = gcd(Math.abs(b[i]), Math.abs(b[i+1]));
			b[i] /= g; b[i+1] /= g;
		}
	}
}