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;
  using ll = long long;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  using vi = vector<int>;
  using vl = vector<ll>;
  using vii = vector<pii>;
  using vll = vector<pll>;
  void debug_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long double eps = 1e-9;
  const long long mod = 1000000007;
  const int MAXN = 200000;
  void solve() {
    ll n, small = INT_MAX, index = 0;
    cin >> n;
    ll arr[n];
    for (int i = 0, nnnn = (n); i < nnnn; ++i) {
      cin >> arr[i];
      if (abs(arr[i]) < small) small = abs(arr[i]), index = i;
    }
    ll b[n];
    memset(b, 0, sizeof b);
    ll t = 0;
    for (int i = 0, nnnn = (n); i < nnnn; ++i) {
      if (i != index) {
        b[i] -= (t / arr[i] / arr[index]) * arr[index];
        if (b[i] == 0) {
          if (t < 0)
            b[i] = arr[index];
          else
            b[i] = -arr[index];
        }
        t += b[i] * arr[i];
      }
    }
    if (t / arr[index] == 0) {
      for (int i = 0, nnnn = (n); i < nnnn; ++i) {
        if (i != index) {
          t -= b[i] * arr[i];
          b[i] += arr[index];
          if (b[i] == 0) b[i] -= 2 * arr[index];
          t += b[i] * arr[i];
          break;
        }
      }
    }
    b[index] = -1 * t / arr[index];
    for (int i = 0, nnnn = (n); i < nnnn; ++i) {
      cout << b[i] << ' ';
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << setprecision(12) << fixed;
    int t;
    cin >> t;
    while (t--) solve();
  }
  