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 long long M = 1e9 + 7;
  void solve() {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; ++i) {
      cin >> a[i];
    }
    if (n % 2 == 0) {
      for (long long i = 0; i < n; i += 2) {
        cout << -a[i + 1] << " " << a[i] << " ";
      }
      cout << '\n';
      return;
    }
    for (long long i = 0; i < n - 3; i += 2) {
      cout << -a[i + 1] << " " << a[i] << " ";
    }
    if (a[n - 1] + a[n - 2] != 0) {
      cout << (a[n - 1] + a[n - 2]) << " " << -a[n - 3] << " " << -a[n - 3]
           << '\n';
      return;
    }
    if (a[n - 3] + a[n - 2] != 0) {
      cout << -a[n - 1] << " " << -a[n - 1] << " " << (a[n - 3] + a[n - 2])
           << '\n';
      return;
    }
    if (a[n - 1] + a[n - 3] != 0) {
      cout << -a[n - 2] << " " << (a[n - 1] + a[n - 3]) << " " << -a[n - 2]
           << '\n';
      return;
    }
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t = 1;
    cin >> t;
    for (long long i = 1; i <= t; i++) {
      solve();
    }
  }
  