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;
  vector<pair<int, int>> solve(int x, int y, int z) {
    vector<pair<int, int>> v = vector<pair<int, int>>{pair<int, int>(abs(x), x),
                                                      pair<int, int>(abs(y), y),
                                                      pair<int, int>(abs(z), z)};
    sort(v.begin(), v.end());
    if (v[2].first % v[0].first != 0) {
      int d = v[2].first / v[0].first, r = v[2].first % v[0].first;
      vector<pair<int, int>> m = vector<pair<int, int>>();
      m.push_back(
          pair<int, int>(v[0].second, d * v[1].first * v[0].second / v[0].first));
      m.push_back(pair<int, int>(v[1].second, r * v[1].second / v[1].first));
      m.push_back(
          pair<int, int>(v[2].second, -v[1].first * v[2].second / v[2].first));
      return m;
    }
    if (v[2].first % v[1].first != 0) {
      int d = v[2].first / v[1].first, r = v[2].first % v[1].first;
      vector<pair<int, int>> m = vector<pair<int, int>>();
      m.push_back(
          pair<int, int>(v[1].second, d * v[0].first * v[1].second / v[1].first));
      m.push_back(pair<int, int>(v[0].second, r * v[0].second / v[0].first));
      m.push_back(
          pair<int, int>(v[2].second, -v[0].first * v[2].second / v[2].first));
      return m;
    } else {
      vector<pair<int, int>> m = vector<pair<int, int>>();
      m.push_back(pair<int, int>(v[2].second, -2));
      m.push_back(pair<int, int>(v[1].second, v[2].second / v[1].second));
      m.push_back(pair<int, int>(v[0].second, v[2].second / v[0].second));
      return m;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t, n, a, b, c;
    cin >> t;
    while (t--) {
      cin >> n;
      if (n % 2) {
        n -= 3;
        cin >> a >> b >> c;
        vector<pair<int, int>> m = solve(a, b, c);
        set<int> s = set<int>{0, 1, 2};
        for (int x : {a, b, c}) {
          for (int i : s) {
            if (m[i].first == x) {
              cout << m[i].second << ' ';
              s.erase(i);
              break;
            }
          }
        }
      }
      while (n) {
        n -= 2;
        cin >> a >> b;
        cout << -b << ' ' << a << ' ';
      }
      cout << '\n';
    }
    return 0;
  }
  