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>
  #pragma GCC optimize(2)
  #pragma GCC optimize(3, "Ofast", "inline")
  using namespace std;
  long long read() {
    long long x = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') ch = getchar();
    while (ch >= '0' && ch <= '9')
      x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
    return x;
  }
  template <typename T>
  inline void rmqmin(T f[][20], T n) {
    for (int i = 1; i <= n; i++) cin >> f[i][0];
    for (int j = 1; j <= 20; j++)
      for (int i = 1; i <= n; i++)
        if (i + (1 << j) - 1 <= n)
          f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
  }
  template <typename T>
  inline T zymin(T f[][20], T z, T y) {
    int x = int(log(y - z + 1) / log(2));
    return min(f[z][x], f[y - (1 << x) + 1][x]);
  }
  template <typename T>
  inline void rmqmax(T f[][20], T n) {
    for (int i = 1; i <= n; i++) cin >> f[i][0];
    for (int j = 1; j <= 20; j++)
      for (int i = 1; i <= n; i++)
        if (i + (1 << j) - 1 <= n)
          f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
  }
  template <typename T>
  inline T zymax(T f[][20], T z, T y) {
    int x = int(log(y - z + 1) / log(2));
    return max(f[z][x], f[y - (1 << x) + 1][x]);
  }
  template <typename T>
  inline void stgcd(T f[][20], T n) {
    for (int i = 1; i <= n; i++) cin >> f[i][0], f[i + n][0] = f[i][0];
    for (int j = 1; j <= 20; j++) {
      for (int i = 1; i + (1 << j) - 1 <= 2 * n; i++) {
        f[i][j] = __gcd(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
      }
    }
  }
  template <typename T>
  inline T qgcd(T f[][20], T z, T y) {
    int x = int(log(y - z + 1) / log(2));
    return __gcd(f[z][x], f[y - (1 << x) + 1][x]);
  }
  long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  const int mod = 998244353;
  int arr[1000005];
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      for (int i = 1; i <= n; i++) cin >> arr[i];
      if (n % 2 == 0) {
        for (int i = 1; i <= n / 2; i++) cout << -arr[n - i + 1] << " ";
        for (int i = n / 2 + 1; i <= n; i++) cout << arr[n - i + 1] << " ";
        cout << endl;
      } else {
        if (arr[1] + arr[2] != 0)
          cout << -arr[3] << " " << -arr[3] << " " << arr[1] + arr[2] << " ";
        else if (arr[1] + arr[3] != 0)
          cout << -arr[2] << " " << arr[1] + arr[3] << " " << -arr[2] << " ";
        else if (arr[2] + arr[3] != 0)
          cout << arr[2] + arr[3] << " " << -arr[1] << " " << -arr[1] << " ";
        for (int i = 4; i <= (n - 3) / 2 + 3; i++) cout << -arr[n - i + 4] << " ";
        for (int i = (n - 3) / 2 + 4; i <= n; i++) cout << arr[n - i + 4] << " ";
        cout << endl;
      }
    }
  }
  