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;
  const long long mod = 1000000007;
  const long long modx = 998244353;
  long double PI = 3.1415926535897;
  const long long N = 200010;
  long long powmod(long long x1, long long y1) {
    long long r1 = 1;
    x1 = x1 % mod;
    while (y1 > 0) {
      if (y1 & 1) r1 = (r1 * x1) % mod;
      y1 = y1 >> 1;
      x1 = (x1 * x1) % mod;
    }
    return r1;
  }
  long long gcd(long long a, long long b) {
    if (b == 0)
      return a;
    else
      return gcd(b, a % b);
  }
  long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  long long ceils(long long x, long long y) { return x / y + ((x % y) != 0); }
  void Unstoppable_1();
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long test = 1;
    cin >> test;
    while (test--) {
      Unstoppable_1();
      cout << "\n";
    }
    return 0;
  }
  void Unstoppable_1() {
    long long n, ans1 = 0, ans2 = 0, ans3 = 0, m = 0, m1 = 0, ans = -1, a1, a2,
                 a3;
    cin >> n;
    vector<long long> v(n), a;
    for (long long i = 0; i < n; i++) cin >> v[i];
    if (n % 2 == 1) {
      a1 = abs(v[0]);
      a2 = abs(v[1]);
      a3 = abs(v[2]);
      ans = lcm(a1, a2);
      ans1 = lcm(a2, a3);
      if (ans > ans1) {
        if (v[0] > 0)
          a.push_back(-(abs(ans / a1)));
        else
          a.push_back((abs(ans / a1)));
        ans3 = ans - ans1;
        if (v[1] < 0)
          a.push_back(-(ans3 / a2));
        else
          a.push_back(ans3 / v[1]);
        if (v[2] < 0)
          a.push_back(-(ans1 / a3));
        else
          a.push_back(ans1 / v[2]);
      } else {
        if (ans == ans1) ans1 *= 2;
        if (v[0] > 0)
          a.push_back((abs(ans / a1)));
        else
          a.push_back(-(abs(ans / a1)));
        ans3 = ans1 - ans;
        if (v[1] < 0)
          a.push_back(-(ans3 / a2));
        else
          a.push_back(ans3 / v[1]);
        if (v[2] < 0)
          a.push_back((ans1 / a3));
        else
          a.push_back(-(ans1 / v[2]));
      }
    }
    if (n % 2 == 0)
      ans = 0;
    else
      ans = 3;
    for (long long i = ans; i < n - 1; i++) {
      if (v[i] > 0)
        a.push_back(abs(v[i + 1]));
      else
        a.push_back(-abs(v[i + 1]));
      if (v[i + 1] > 0)
        a.push_back(-abs(v[i]));
      else
        a.push_back(abs(v[i]));
      i += 1;
    }
    for (long long i = 0; i < n; i++) cout << a[i] << " ";
  }
  