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("Ofast")
  using namespace std;
  template <typename T>
  long long powe(T base, T to) {
    long long ans = 1;
    while (to > 0) {
      if (to & 1)
        ans = ((ans) * (base)), to--;
      else
        base = ((base) * (base)), to /= 2;
    }
    return ans;
  }
  long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
  bool prime(long long n) {
    if (n == 1) return 0;
    for (long long i = 2; i * i <= n; i++)
      if (n % i == 0) return 0;
    return 1;
  }
  int binarySearch(int arr[], int start, int end, int val) {
    int mid;
    end--;
    while (start <= end) {
      mid = start + (end - start) / 2;
      if (arr[mid] == val)
        return mid;
      else if (arr[mid] < val)
        start = mid + 1;
      else
        end = mid - 1;
    }
    return -1;
  }
  bool check(string str) {
    bool ans = true;
    for (int i = 0; i < str.size() / 2; i++)
      if (str[i] != str[str.size() - i - 1]) ans = false;
    return ans;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int tl = 1;
    cin >> tl;
    while (tl--) {
      int n;
      cin >> n;
      int arr[n];
      for (auto &i : arr) cin >> i;
      ;
      int end = n;
      if (n & 1) end = n - 3;
      vector<int> b(n, 0);
      for (int i = 0; i < end; i += 2) {
        b[i] = -arr[i + 1];
        b[i + 1] = arr[i];
      }
      if (end != n) {
        if ((arr[n - 2] + arr[n - 3]) != 0)
          b[n - 1] = arr[n - 2] + arr[n - 3], b[n - 2] = -arr[n - 1],
                b[n - 3] = -arr[n - 1];
        else if ((arr[n - 2] + arr[n - 1]) != 0)
          b[n - 3] = arr[n - 2] + arr[n - 1], b[n - 2] = -arr[n - 3],
                b[n - 1] = -arr[n - 3];
        else
          b[n - 2] = arr[n - 3] + arr[n - 1], b[n - 3] = -arr[n - 2],
                b[n - 1] = -arr[n - 2];
      }
      for (int i : b) cout << i << " ";
      cout << "\n";
    }
  }
  