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.*;
import java.io.*;

public class MyClass {
    
    public static void main(String args[]) throws Exception {
      BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
      
      int T = Integer.parseInt(s.readLine());
      
      for (int t = 0; t < T; ++t) {
          int n = Integer.parseInt(s.readLine());
          
          int[] a = new int[n + 5];
          int[] b = new int[n + 5];
          String[] ss = s.readLine().split(" ");
          
          for (int i = 0; i < n; ++i) {
              a[i] = Integer.parseInt(ss[i]);
          }
          
          if (n % 2 == 1) {
              int Q = solve(a[n - 3], a[n - 2] + a[n - 1]);
              for (int k = 1; k <= 3; ++k) {
                  int c = a[n - k];
                  int d = 0;
                  for (int p = 1; p <= 3; ++p) if (p != k) {
                      d += a[n - p];
                  }
                  
                  if (c == 0 || d == 0) {
                      continue;
                  }
                  
                  Q = solve(c, d);
                  for (int p = 1; p <= 3; ++p) if (p != k) {
                      b[n - p] = -Q / d;
                  } else {
                      b[n - p] = Q / c;
                  }
              }
              
              
              for (int i = 0; i < n - 3; i += 2) {
                  Q = solve(a[i], a[i + 1]);
                  b[i] = Q / a[i];
                  b[i + 1] = -Q / a[i + 1];
              }
          } else {
             for (int i = 0; i < n; i += 2) {
                  int Q = solve(a[i], a[i + 1]);
                  b[i] = Q / a[i];
                  b[i + 1] = -Q / a[i + 1];
              }  
          }
          
          for (int i = 0; i < n - 1; ++i) {
              System.out.print(b[i] + " ");
          }
          System.out.println(b[n - 1]);
      }
    }
    
    private static int solve(int a, int b) {
        return a * b / gcd(a, b);
    }
    
    private static int gcd(int a, int b) {
        return (b == 0) ? a : gcd(b, a % b);
    }
}