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>
  const int N = 1e5 + 5;
  const int mod = 1e9 + 7;
  const long long int int_max = 1e18;
  const long long int int_min = -1e18;
  const long double PI = acos(-1);
  using namespace std;
  long long int power(long long int x, long long int y) {
    long long int temp;
    if (y == 0) return 1;
    temp = power(x, y / 2);
    if (y % 2 == 0)
      return temp * temp;
    else {
      if (y > 0)
        return x * temp * temp;
      else
        return (temp * temp) / x;
    }
  }
  void solve() {
    long long int n;
    cin >> n;
    vector<pair<long long int, long long int>> v;
    vector<long long int> ckc(n, 0);
    long long int a1[n + 2];
    for (long long int i = 0; i <= n - 1; i++) {
      cin >> a1[i];
      v.push_back({a1[i], i});
    }
    if (n % 2 == 0) {
      vector<long long int> finalans;
      for (long long int i = 1; i <= n; i++) {
        finalans.push_back(-v[i].first);
        finalans.push_back(v[i - 1].first);
        i++;
      }
      for (long long int i : finalans) {
        cout << i << ' ';
      }
      cout << '\n';
      return;
    }
    long long int flg = 0;
    long long int idx = -1;
    for (long long int i = 0; i <= n - 1; i++) {
      if (a1[i] == 1) {
        idx = i;
        flg = 1;
        break;
      }
    }
    long long int sum = 0;
    sort(v.begin(), v.end());
    long long int b[n + 2], a[n + 2];
    for (int i = 0; i <= n - 1; i++) {
      b[i] = v[i].first;
    }
    for (int i = 0; i < n; i++) {
      a[i] = v[i].second;
    }
    if ((b[0] + b[1]) != 0) {
      ckc[a[0]] = -b[2];
      ckc[a[1]] = -b[2];
      ckc[a[2]] = b[0] + b[1];
    } else if ((b[1] + b[2]) != 0) {
      ckc[a[1]] = -b[0];
      ckc[a[0]] = b[2] + b[1];
      ckc[a[2]] = -b[0];
    } else if ((b[2] + b[0]) != 0) {
      ckc[a[2]] = -b[1];
      ckc[a[1]] = b[0] + b[2];
      ckc[a[0]] = -b[1];
    }
    for (long long int i = 4; i <= n - 1; i++) {
      ckc[a[i]] = b[i - 1];
      ckc[a[i - 1]] = -b[i];
      i++;
    }
    for (long long int i = 0; i <= n - 1; i++) {
      cout << ckc[i] << ' ';
    }
    cout << '\n';
  }
  int main() {
    ios::sync_with_stdio(false), cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  