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 ll = long long;
  using ld = long double;
  using Pi = pair<ll, ll>;
  using Tup = tuple<ll, ll, ll>;
  using vl = vector<ll>;
  using vd = vector<ld>;
  using vvd = vector<vd>;
  using vc = vector<char>;
  using vb = vector<bool>;
  using vs = vector<string>;
  using vp = vector<Pi>;
  using vvc = vector<vector<char>>;
  using vvl = vector<vector<ll>>;
  using vvvl = vector<vector<vector<ll>>>;
  using vvvvl = vector<vvvl>;
  const ll INF = 1LL << 60;
  const ld PI = 3.1415926535897932385;
  template <class T>
  bool chmax(T &a, const T &b) {
    if (a < b) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  bool chmin(T &a, const T &b) {
    if (b < a) {
      a = b;
      return 1;
    }
    return 0;
  }
  ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
  ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
  void printb(ll N, ll d = 16) {
    for (ll i = 0; i < (d); ++i) {
      cout << (N / ((1LL) << (d - i - 1))) % 2;
    }
    cout << endl;
  }
  void printv(vector<ll> a) {
    for (ll i = 0; i < (a.size()); ++i) {
      if (i == a.size() - 1) {
        cout << a[i] << endl;
      } else {
        cout << a[i] << " ";
      }
    }
  }
  void printv2(vector<ll> a) {
    for (ll i = 0; i < (a.size()); ++i) {
      cout << a[i] << endl;
    }
  }
  void printvv(vvl a) {
    for (ll i = 0; i < (a.size()); ++i) {
      printv(a[i]);
    }
  }
  bool In_map(ll y, ll x, ll h, ll w) {
    if (y < 0 || x < 0 || y >= h || x >= w) {
      return 0;
    } else {
      return 1;
    }
  }
  bool compare(Pi a, Pi b) {
    if (a.first != b.first) {
      return a.first < b.first;
    } else {
      return a.second < b.second;
    }
  }
  const vector<ll> dx{1, 0, -1, 0};
  const vector<ll> dy{0, 1, 0, -1};
  void Main() {
    ll n;
    cin >> n;
    vl a(n);
    map<ll, ll> m;
    vl b(n);
    ll flag = 0;
    for (ll i = 0; i < (n); ++i) {
      cin >> a[i];
    }
    if (n % 2 == 1) {
      if (a[0] == -a[1]) {
        if (a[0] == -a[2]) {
          b[0] = -(a[1] + a[2]);
          b[1] = a[0];
          b[2] = a[0];
        } else {
          b[0] = a[1];
          b[1] = -(a[0] + a[2]);
          b[2] = a[1];
        }
      } else {
        b[0] = a[2];
        b[1] = a[2];
        b[2] = -(a[0] + a[1]);
      }
      ll p = 0;
      ll pi;
      for (ll i = 3; i < (n); ++i) {
        if (b[i] != 0) continue;
        if (p != 0) {
          b[i] = p;
          b[pi] = -a[i];
          p = 0;
        } else {
          p = a[i];
          pi = i;
        }
      }
    } else {
      ll p = 0;
      ll pi;
      for (ll i = 0; i < (n); ++i) {
        if (b[i] != 0) continue;
        if (p != 0) {
          b[i] = p;
          b[pi] = -a[i];
          p = 0;
        } else {
          p = a[i];
          pi = i;
        }
      }
    }
    printv(b);
    return;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
    cin >> t;
    while (t--) {
      Main();
    }
    return 0;
  }
  