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 LLmn = LLONG_MIN;
  const long long LLmx = LLONG_MAX;
  const long long MOD = 1e6 + 3;
  const long long N = 1e9 + 7;
  const long double PI = 3.14159265358979323846;
  long long gcd(long long a, long long b) {
    if (a == 0) {
      return b;
    }
    return gcd(b % a, a);
  }
  long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
  pair<int, int> w[9] = {make_pair(-1, 0), make_pair(0, -1),  make_pair(1, 0),
                         make_pair(0, 1),  make_pair(-1, -1), make_pair(0, 1),
                         make_pair(1, -1), make_pair(1, 1)};
  void solve() {
    long long n;
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    if (n & 1) {
      if (a[0] + a[1] != 0)
        cout << -a[2] << " " << -a[2] << " " << a[0] + a[1] << " ";
      else if (a[0] + a[2] != 0)
        cout << -a[1] << " " << a[0] + a[2] << " " << -a[1] << " ";
      else
        cout << a[1] + a[2] << " " << -a[0] << " " << -a[0] << " ";
      for (long long i = 3; i < n; i += 2) {
        if (i + 1 < n) cout << a[i + 1] << " " << -a[i] << " ";
      }
      cout << "\n";
      ;
    } else {
      for (long long i = 0; i < n; i += 2) {
        if (i + 1 < n) cout << a[i + 1] << " " << -a[i] << " ";
      }
      cout << "\n";
      ;
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    ;
    long long t;
    cin >> t;
    while (t--) solve();
  }
  