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;
  int f(int a, long long int* d) {
    if (a == d[a]) return a;
    return d[a] = f(d[a], d);
  }
  void u(int a, int b, long long int* d, long long int* s) {
    a = f(a, d);
    b = f(b, d);
    if (a != b) {
      d[b] = a;
      s[a] += s[b];
    }
  }
  int gcd(int a, int b) {
    while (b) {
      a %= b;
      swap(a, b);
    }
    return a;
  }
  long long int binpow(long long int a, long long int b) {
    long long int res = 1;
    while (b > 0) {
      if (b & 1) res = res * a;
      a = a * a;
      b >>= 1;
    }
    return res;
  }
  int main() {
    int w;
    long long int a, b, n, c;
    long long int t;
    cin >> w;
    while (w--) {
      cin >> n;
      int d[n + 1];
      for (int i = 0; i < n; i++) cin >> d[i];
      int p[n + 1];
      if (n % 2 == 0) {
        for (int i = 0; i < n; i++) {
          if (i % 2 == 0)
            p[i] = -d[i + 1];
          else
            p[i] = d[i - 1];
        }
      } else {
        for (int i = 0; i < n - 3; i++) {
          if (i % 2 == 0)
            p[i] = -d[i + 1];
          else
            p[i] = d[i - 1];
        }
        int f = 0;
        for (int i = n - 3; i <= n - 1; i++) {
          for (int j = n - 3; j <= n - 1; j++) {
            if (i != j && (d[i] + d[j]) != 0) {
              int k;
              for (k = n - 3; k <= n - 1; k++) {
                if (k != i && k != j) break;
              }
              p[k] = -(d[i] + d[j]);
              p[i] = d[k];
              p[j] = d[k];
              f = 1;
              break;
            }
          }
          if (f) break;
        }
      }
      for (int i = 0; i < n; i++) cout << p[i] << " ";
      cout << "\n";
    }
  }
  