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;
  namespace {
  using T = long long;
  namespace module {
  inline const T m = 998244353;
  void add(T& a, T b) {
    if ((a += b) >= m) a -= m;
  }
  void dec(T& a, T b) {
    if ((a -= b) < 0) a += m;
  }
  void mul(T& a, T b) { a = a * b % m; }
  T radd(T a, T b) { return (a += b) >= m ? a - m : a; }
  T rdec(T a, T b) { return (a -= b) < 0 ? a + m : a; }
  T rmul(T a, T b) { return a * b % m; }
  };  // namespace module
  constexpr char endl = '\n';
  const int maxN = 2e5 + 5;
  int n;
  T a[maxN], b[maxN];
  T gcd(T n, T m) { return m ? gcd(m, n % m) : n; }
  void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i += 2) {
      b[i] = a[i + 1];
      b[i + 1] = -a[i];
      T g = gcd(abs(b[i]), abs(b[i + 1]));
      b[i] /= g;
      b[i + 1] /= g;
    }
    if (n & 1) {
      int i = n - 2, j = n - 1, k = n;
      if (a[j] + a[k] == 0) i = n - 1, j = n - 2, k = n;
      if (a[j] + a[k] == 0) i = n, j = n - 2, k = n - 1;
      b[i] = a[j] + a[k];
      b[j] = -a[i];
      b[k] = -a[i];
      T g = gcd(abs(b[i]), abs(b[j]));
      g = gcd(g, abs(b[k]));
      b[i] /= g;
      b[j] /= g;
      b[k] /= g;
    }
    T sum = 0;
    for (int i = 1; i <= n; i++) sum += a[i] * b[i];
    if (sum) cerr << "NO" << endl;
    for (int i = 1; i <= n; i++) sum += abs(b[i]);
    assert(sum <= 1000000000);
    for (int i = 1; i <= n; i++) cout << b[i] << " \n"[i == n];
  }
  }  // namespace
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    int t(1);
    cin >> t;
    for (int i = 1; i <= t; i++) solve();
  }
  