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>
  #pragma GCC optimize("O4,unroll-loops,no-stack-protector")
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  void Why_does_competitive_programming_even_exist() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  }
  void YayXD() { return; }
  pair<long long, long long> arr[100005];
  int ans[100005];
  bool comp(pair<long long, long long> a, pair<long long, long long> b) {
    return abs(a.first) < abs(b.first);
  }
  long long gcd(long long a, long long b) {
    if (a == 0) return b;
    if (b == 0) return a;
    return gcd(b % a, a);
  }
  signed main() {
    Why_does_competitive_programming_even_exist();
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      for (long long i = 1; i <= n; i++) {
        cin >> arr[i].first;
        arr[i].second = i;
      }
      if (n % 2 == 0) {
        for (long long i = 1; i <= n; i += 2) {
          cout << arr[i + 1].first << " " << -arr[i].first << " ";
        }
        cout << "\n";
        continue;
      }
      sort(arr + 1, arr + 1 + n, comp);
      long long a = abs(arr[1].first), b = abs(arr[2].first),
                c = abs(arr[3].first), g = gcd(a, gcd(b, c));
      a /= g;
      b /= g;
      c /= g;
      if (b == c) {
        long long x = b * 2, y = a;
        if (arr[1].first > 0)
          ans[arr[1].second] = -x;
        else
          ans[arr[1].second] = x;
        if (arr[2].first > 0)
          ans[arr[2].second] = y;
        else
          ans[arr[2].second] = -y;
        if (arr[3].first > 0)
          ans[arr[3].second] = y;
        else
          ans[arr[3].second] = -y;
      } else {
        long long y = a, x = c - b;
        if (arr[1].first > 0)
          ans[arr[1].second] = -x;
        else
          ans[arr[1].second] = x;
        if (arr[2].first > 0)
          ans[arr[2].second] = -y;
        else
          ans[arr[2].second] = y;
        if (arr[3].first > 0)
          ans[arr[3].second] = y;
        else
          ans[arr[3].second] = -y;
      }
      for (long long i = 4; i <= n; i += 2) {
        ans[arr[i].second] = -arr[i + 1].first;
        ans[arr[i + 1].second] = arr[i].first;
      }
      for (int i = 1; i <= n; i++) cout << ans[i] << " ";
      cout << "\n";
    }
  }
  