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;
  const long long maxi = 1000001;
  const long long inf = 1e16;
  const long long ninf = 1e16 * (-1);
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long power(long long a, long long b) {
    if (b == 0) return 1;
    if (b == 1) return a;
    long long ret = (power(a, b / 2)) % 1000000007;
    if (b % 2 == 0) return (ret * ret) % 1000000007;
    return ((ret * ret) % 1000000007 * a) % 1000000007;
  }
  long long power1(long long a, long long b) {
    if (b == 0) return 1;
    if (b == 1) return a;
    long long ret = (power(a, b / 2));
    if (b % 2 == 0) return (ret * ret);
    return ((ret * ret) * a);
  }
  long long modInverse(long long n) { return power(n, 1000000007 - 2); }
  long long vis[maxi] = {0};
  signed main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    ;
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      long long a[n];
      long long b[n];
      for (long long i = 0; i < n; i++) {
        cin >> a[i];
        b[i] = abs(a[i]);
      }
      vector<long long> v;
      long long z;
      if (n % 2 == 0) {
        for (long long i = 0; i < n; i += 2) {
          z = (b[i] * b[i + 1]) / gcd(b[i], b[i + 1]);
          if ((a[i] > 0 && a[i + 1] < 0) || a[i] < 0 && a[i + 1] > 0) {
            v.push_back(z / b[i]);
            v.push_back(z / b[i + 1]);
          } else {
            v.push_back(z / b[i]);
            v.push_back(0 - (z / b[i + 1]));
          }
        }
      } else {
        for (long long i = 0; i < n - 3; i += 2) {
          z = (b[i] * b[i + 1]) / gcd(b[i], b[i + 1]);
          if ((a[i] > 0 && a[i + 1] < 0) || a[i] < 0 && a[i + 1] > 0) {
            v.push_back(z / b[i]);
            v.push_back(z / b[i + 1]);
          } else {
            v.push_back(z / b[i]);
            v.push_back(0 - (z / b[i + 1]));
          }
        }
        long long ansh = a[n - 3] + a[n - 2];
        z = (b[n - 1] * abs(ansh)) / gcd(abs(ansh), b[n - 1]);
        if (ansh == 0 && a[n - 3] > 0) {
          v.push_back(b[n - 1]);
          v.push_back(0 - b[n - 1]);
          if (a[n - 1] > 0)
            v.push_back(0 - (b[n - 3] + b[n - 2]));
          else
            v.push_back((b[n - 3] + b[n - 2]));
        } else if (ansh == 0 && a[n - 3] < 0) {
          v.push_back(0 - b[n - 1]);
          v.push_back(b[n - 1]);
          if (a[n - 1] > 0)
            v.push_back(0 - (b[n - 3] + b[n - 2]));
          else
            v.push_back((b[n - 3] + b[n - 2]));
        } else if ((ansh > 0 && a[n - 1] < 0) || (ansh < 0 && a[n - 1] > 0)) {
          v.push_back(z / abs(ansh));
          v.push_back(z / abs(ansh));
          v.push_back(z / b[n - 1]);
        } else {
          v.push_back(0 - (z / abs(ansh)));
          v.push_back(0 - (z / abs(ansh)));
          v.push_back(z / b[n - 1]);
        }
      }
      for (long long i = 0; i < v.size(); i++) {
        cout << v[i] << " ";
      }
      cout << endl;
    }
    return 0;
  }
  