description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int N = 1e5 + 10;
  int a[N];
  int n;
  int res[N];
  int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
  int lcm(int a, int b) { return a * b / gcd(a, b); }
  void solve() {
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    if (n % 2 == 0) {
      for (int i = 0; i < n / 2; i++) {
        int j = n - 1 - i;
        int t = lcm(a[i], a[j]);
        res[i] = t / a[i];
        res[j] = t / a[j];
        if (1) {
          res[i] *= -1;
        }
      }
      for (int i = 0; i < n; i++) cout << res[i] << " ";
      cout << endl;
    } else {
      for (int i = 0; i < n / 2 - 1; i++) {
        int j = n - 1 - i;
        int t = lcm(a[i], a[j]);
        res[i] = t / a[i];
        res[j] = t / a[j];
        if (1) {
          res[i] *= -1;
        }
      }
      res[n / 2 - 1] = 1;
      res[n / 2 + 1] = -1;
      int t = 1 * a[n / 2 - 1] + -1 * a[n / 2 + 1];
      if (t == 0) {
        t += a[n / 2 - 1];
        res[n / 2 - 1]++;
      }
      int t2 = lcm(t, a[n / 2]);
      res[n / 2] = t2 / a[n / 2];
      res[n / 2 - 1] *= t2 / t;
      res[n / 2 + 1] *= t2 / t;
      if (1) {
        res[n / 2] *= -1;
      }
      for (int i = 0; i < n; i++) cout << res[i] << " ";
      cout << endl;
    }
  }
  int main() {
    int K;
    scanf("%d", &K);
    while (K--) {
      solve();
    }
  }
  