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 static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;
/*
-> Give your 100%, that's it!
-> Rules To Solve Any Problem:
    1. Read the problem.
    2. Think About It.
    3. Solve it!
*/

public class Template {
    
	static int mod = 1000000007;

	public static void main(String[] args){
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int yo = sc.nextInt();
        while (yo-- > 0) {
            int n = sc.nextInt();
            int[] a = sc.readInts(n);
            
            if(n%2 == 0){
                int[] b = new int[n];
                for(int i = 0; i < n; i+=2){
                    b[i] = -a[i+1];
                    b[i+1] = a[i];
                }
                // int sum = 0;
                // for(int i = 0; i < n; i++){
                //     sum += a[i]*b[i];
                // }
                // out.println(sum);
                print(b,out);
                continue;
            }
            
            int[] b = new int[n];
            
            int a1 = a[0];
            int a2 = a[1];
            int a3 = a[2];
            
            if(a1 + a2 != 0){
                b[0] = a3;
                b[1] = a3;
                b[2] = -a1-a2;
            }
            else {
                b[0] = a3;
                b[1] = -a3;
                b[2] = 2*a1;
            }
            
            for(int i = 3; i < n; i+=2){
                b[i] = -a[i+1];
                b[i+1] = a[i];
            }
            // int sum = 0;
            // for(int i = 0; i < n; i++){
            //     sum += a[i]*b[i];
            // }
            // out.println(sum);
            print(b,out);
        }
        out.close();
	}
    
   
    
    
    /*
    Source: hu_tao
         Random stuff to try when stuck:
            - use bruteforcer
            - always check for n = 1, n = 2, so on
            -if it's 2C then it's dp
            -for combo/probability problems, expand the given form we're interested in
            -make everything the same then build an answer (constructive, make everything 0 then do something)
            -something appears in parts of 2 --> model as graph
            -assume a greedy then try to show why it works
            -find way to simplify into one variable if multiple exist
            -treat it like fmc (note any passing thoughts/algo that could be used so you can revisit them)
            -find lower and upper bounds on answer
            -figure out what ur trying to find and isolate it
            -see what observations you have and come up with more continuations
            -work backwards (in constructive, go from the goal to the start)
            -turn into prefix/suffix sum argument (often works if problem revolves around adjacent array elements)
            -instead of solving for answer, try solving for complement (ex, find n-(min) instead of max)
            -draw something
            -simulate a process
            -dont implement something unless if ur fairly confident its correct
            -after 3 bad submissions move on to next problem if applicable
            -do something instead of nothing and stay organized
            -write stuff down
         Random stuff to check when wa:
            -if code is way too long/cancer then reassess
            -switched N/M
            -int overflow
            -switched variables
            -wrong MOD
            -hardcoded edge case incorrectly
         Random stuff to check when tle:
            -continue instead of break
            -condition in for/while loop bad
         Random stuff to check when rte:
            -switched N/M
            -long to int/int overflow
            -division by 0
            -edge case for empty list/data structure/N=1
      */

	public static class Pair {
		int x;
		int y;

		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
	}

	public static void sort(int[] arr) {
		ArrayList<Integer> ls = new ArrayList<Integer>();
		for (int x : arr)
			ls.add(x);
		Collections.sort(ls);
		for (int i = 0; i < arr.length; i++)
			arr[i] = ls.get(i);
	}

	public static long gcd(long a, long b) {
		if (b == 0)
			return a;
		return gcd(b, a % b);
	}

	static boolean[] sieve(int N) {
		boolean[] sieve = new boolean[N + 1];
		for (int i = 2; i <= N; i++)
			sieve[i] = true;

		for (int i = 2; i <= N; i++) {
			if (sieve[i]) {
				for (int j = 2 * i; j <= N; j += i) {
					sieve[j] = false;
				}
			}
		}
		return sieve;
	}

	public static long power(long x, long y, long p) {
		long res = 1L;
		x = x % p;
		while (y > 0) {
			if ((y & 1) == 1)
				res = (res * x) % p;
			y >>= 1;
			x = (x * x) % p;
		}
		return res;
	}

	public static void print(int[] arr, PrintWriter out) {
		//for debugging only
		for (int x : arr)
			out.print(x + " ");
		out.println();
	}
    
    public static int log2(int a){
        return (int)(Math.log(a)/Math.log(2));
    }

	static class FastScanner {
		private int BS = 1 << 16;
		private char NC = (char) 0;
		private byte[] buf = new byte[BS];
		private int bId = 0, size = 0;
		private char c = NC;
		private double cnt = 1;
		private BufferedInputStream in;

		public FastScanner() {
			in = new BufferedInputStream(System.in, BS);
		}

		public FastScanner(String s) {
			try {
				in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
			} catch (Exception e) {
				in = new BufferedInputStream(System.in, BS);
			}
		}

		private char getChar() {
			while (bId == size) {
				try {
					size = in.read(buf);
				} catch (Exception e) {
					return NC;
				}
				if (size == -1)
					return NC;
				bId = 0;
			}
			return (char) buf[bId++];
		}

		public int nextInt() {
			return (int) nextLong();
		}

		public int[] readInts(int N) {
			int[] res = new int[N];
			for (int i = 0; i < N; i++) {
				res[i] = (int) nextLong();
			}
			return res;
		}

		public long[] readLongs(int N) {
			long[] res = new long[N];
			for (int i = 0; i < N; i++) {
				res[i] = nextLong();
			}
			return res;
		}

		public long nextLong() {
			cnt = 1;
			boolean neg = false;
			if (c == NC)
				c = getChar();
			for (; (c < '0' || c > '9'); c = getChar()) {
				if (c == '-')
					neg = true;
			}
			long res = 0;
			for (; c >= '0' && c <= '9'; c = getChar()) {
				res = (res << 3) + (res << 1) + c - '0';
				cnt *= 10;
			}
			return neg ? -res : res;
		}

		public double nextDouble() {
			double cur = nextLong();
			return c != '.' ? cur : cur + nextLong() / cnt;
		}

		public double[] readDoubles(int N) {
			double[] res = new double[N];
			for (int i = 0; i < N; i++) {
				res[i] = nextDouble();
			}
			return res;
		}

		public String next() {
			StringBuilder res = new StringBuilder();
			while (c <= 32)
				c = getChar();
			while (c > 32) {
				res.append(c);
				c = getChar();
			}
			return res.toString();
		}

		public String nextLine() {
			StringBuilder res = new StringBuilder();
			while (c <= 32)
				c = getChar();
			while (c != '\n') {
				res.append(c);
				c = getChar();
			}
			return res.toString();
		}

		public boolean hasNext() {
			if (c > 32)
				return true;
			while (true) {
				c = getChar();
				if (c == NC)
					return false;
				else if (c > 32)
					return true;
			}
		}
	}

	//	For Input.txt and Output.txt	
	//	FileInputStream in = new FileInputStream("input.txt");
	//	FileOutputStream out = new FileOutputStream("output.txt");
	//	PrintWriter pw = new PrintWriter(out);
	//	Scanner sc = new Scanner(in);
}