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;
  using namespace chrono;
  void _print(long long int t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{ ";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  long long int mod_add(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long int mod_mul(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long int mod_sub(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long int gcd(long long int a, long long int b) {
    if (b > a) {
      return gcd(b, a);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long int expo(long long int a, long long int b, long long int mod) {
    long long int res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b = b >> 1;
    }
    return res;
  }
  long long int lcm(long long int a, long long int b) {
    return (a / gcd(a, b)) * b;
  }
  void solve() {
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      long long int a[n];
      for (int i = 0; i < n; i++) {
        cin >> a[i];
      }
      long long int b[n];
      int st = 0, end = n - 1;
      while (st < end) {
        b[st] = -a[end];
        b[end] = a[st];
        st++;
        end--;
      }
      long long int temp = INT_MAX;
      long long int ind, val, mid;
      if (n % 2) {
        for (int i = 0; i < n; i++) {
          if (i != (n / 2)) {
            long long int lc = abs((a[i] * a[n / 2]));
            long long int g = gcd(abs(a[n / 2]), abs(a[i]));
            lc /= g;
            long long int x = lc / a[i];
            long long int y = lc / a[n / 2];
            if (x + y < temp) {
              ind = i;
              if (b[i] + x == 0) {
                val = -x;
                mid = y;
              } else {
                val = x;
                mid = -y;
              }
            }
          }
        }
        b[ind] += val;
        b[n / 2] = mid;
      }
      for (int i = 0; i < n; i++) {
        cout << b[i] << " ";
      }
      cout << endl;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    auto start1 = high_resolution_clock::now();
    solve();
    auto stop1 = high_resolution_clock::now();
    auto duration = duration_cast<microseconds>(stop1 - start1);
  }
  