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;
  void init_code() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  const double EPSILON = 10e-8;
  bool equal(double a, double b) { return fabs(a - b) < EPSILON; }
  bool less(double a, double b) { return a < b && !equal(a, b); }
  long long fast_power(long long base, long long power, long long MOD) {
    if (power == 0) return 1;
    if (power == 1) return base % MOD;
    long long result = fast_power(base, power / 2, MOD) % MOD;
    if (power % 2 == 0) return ((result % MOD) * (result % MOD)) % MOD;
    long long a = ((result % MOD) * (base % MOD)) % MOD;
    a = ((a % MOD) * (result % MOD)) % MOD;
    return a % MOD;
  }
  long long fact[1000000];
  void factorial() {
    fact[0] = fact[1] = 1;
    for (long long i = 2; i < 1000000; i++) {
      fact[i] = ((i % 998244353) * (fact[i - 1] % 998244353)) % 998244353;
    }
  }
  int main() {
    init_code();
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      long long arr[n];
      for (long long i = 0; i < n; i++) {
        cin >> arr[i];
      }
      long long b[n];
      if (n % 2 == 0) {
        for (long long i = 0; i < n - 1; i += 2) {
          b[i] = -(arr[i + 1]);
          b[i + 1] = arr[i];
        }
        for (long long i = 0; i < n; i++) {
          cout << b[i] << " ";
        }
        cout << endl;
      } else {
        for (long long i = 0; i < n - 3; i += 2) {
          b[i] = -(arr[i + 1]);
          b[i + 1] = arr[i];
        }
        if (arr[n - 3] + arr[n - 2] != 0) {
          b[n - 3] = b[n - 2] = -arr[n - 1];
          b[n - 1] = arr[n - 3] + arr[n - 2];
        } else if (arr[n - 3] + arr[n - 1] != 0) {
          b[n - 3] = b[n - 1] = -arr[n - 2];
          b[n - 2] = arr[n - 3] + arr[n - 1];
        } else if (arr[n - 2] + arr[n - 1] != 0) {
          b[n - 1] = b[n - 2] = -arr[n - 3];
          b[n - 3] = arr[n - 2] + arr[n - 1];
        }
        for (long long i = 0; i < n; i++) {
          cout << b[i] << " ";
        }
        cout << endl;
      }
    }
  }
  