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 {
    static long gcd(long a, long b) {
        while (b != 0) {
            long buf = a;
            a = b;
            b = buf % b;
        }
        return a;
    }

    public static void main(String[] args) throws IOException {
        int t = nextInt();
        while (t-- != 0) {
            int n = nextInt();
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextLong();
            }
            long[] ans = new long[n];
            if (n % 2 == 0) {
                for (int i = 0; i < n; i += 2) {
                    long x = Math.abs(a[i]);
                    long y = Math.abs(a[i + 1]);
                    long lcm = x * y / gcd(x, y);
                    long d1 = lcm / x;
                    long d2 = lcm / y;
                    if ((a[i] <= 0 && a[i + 1] <= 0) || (a[i] > 0 && a[i + 1] > 0)) {
                        ans[i] = d1;
                        ans[i + 1] = -d2;
                    } else {
                        ans[i] = d1;
                        ans[i + 1] = d2;
                    }
                }
            }
            else{
                for (int i = 0; i < n-2; i += 2) {
                    long x = Math.abs(a[i]);
                    long y = Math.abs(a[i + 1]);
                    long lcm = x * y / gcd(x, y);
                    long d1 = lcm / x;
                    long d2 = lcm / y;
                    if ((a[i] <= 0 && a[i + 1] <= 0) || (a[i] > 0 && a[i + 1] > 0)) {
                        ans[i] = d1;
                        ans[i + 1] = -d2;
                    } else {
                        ans[i] = d1;
                        ans[i + 1] = d2;
                    }
                }
                long x1=Math.abs(a[n-3]);
                long x2=Math.abs(a[n-2]);
                long x3=Math.abs(a[n-1]);
                long l=x1*x2/gcd(x1,x2);
                long r=l*x3/gcd(l,x3);
                ans[n-3]=r/x1;
                ans[n-2]=r/x2;
                if(a[n-3]<=0)ans[n-3]=-r/x1;
                if(a[n-2]<=0)ans[n-2]=-r/x2;
                long q=a[n-3]*ans[n-3]+a[n-2]*ans[n-2];
                if(a[n-1]<=0)ans[n-1]=q/x3;
                else ans[n-1]=-q/x3;
            }
            for (int i = 0; i < n; i++) {
                out.print(ans[i]+" ");
            }
            out.println();
        }
        out.close();
    }

    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer in = new StringTokenizer("");


    public static boolean hasNext() throws IOException {
        if (in.hasMoreTokens()) return true;
        String s;
        while ((s = br.readLine()) != null) {
            in = new StringTokenizer(s);
            if (in.hasMoreTokens()) return true;
        }
        return false;
    }

    public static String nextToken() throws IOException {
        while (!in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public static int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public static long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public static double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}

class Ayo {
    int v;
    long w;

    public Ayo(int v, long w) {
        this.v = v;
        this.w = w;
    }
}

class Ayo1 implements Comparable<Ayo1> {
    long w;
    int v;

    public Ayo1(long w, int v) {
        this.w = w;
        this.v = v;
    }

    @Override
    public int compareTo(Ayo1 o) {
        if (o.v == v && o.w == w) return 0;
        else {
            if (w < o.w) return -1;
            else if (w > o.w) return 1;
            else {
                return Integer.compare(v, o.v);
            }
        }
    }
}