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 M = 1e9 + 7;
  signed long long power_binaryexp(long long a, long long b) {
    long long ans = 1;
    while (b > 0) {
      if (b & 1) {
        ans = (ans * a);
      }
      a = (a * a);
      b = b >> 1;
    }
    return ans;
  }
  signed long long binarymul(long long a, long long b) {
    long long ans = 0;
    while (b) {
      if (b & 1) {
        ans = ans + a;
      }
      a += a;
      b >>= 1;
    }
    return ans;
  }
  bool is_prime(long long n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (long long i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  bool compare(pair<long long, long long> p1, pair<long long, long long> p2) {
    return p1.second > p2.second;
  }
  bool checkp(string s) {
    long long n = s.length();
    for (long long i = 0; i < n / 2; i++) {
      if (s[i] != s[n - i - 1]) return false;
    }
    return true;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long ttt;
    cin >> ttt;
    while (ttt--) {
      long long n;
      cin >> n;
      vector<long long> v(n);
      long long sum = 0;
      long long s = 0;
      for (long long i = 0; i < n; i++) {
        cin >> v[i];
      }
      vector<long long> v1(n);
      if (n & 1) {
        if (v[1] < 0)
          v1[1] = -abs(v[0]);
        else
          v1[1] = abs(v[0]);
        if (v[2] < 0)
          v1[2] = -abs(v[0]);
        else
          v1[2] = abs(v[0]);
        s = (v[1] * v1[1] + v1[2] * v[2]);
        if (v[0] < 0) {
          s = -s;
          v[0] = -v[0];
        }
        v1[0] = -s / v[0];
        if (v[0] < 0) n = -n;
        for (long long i = 3; i < n; i += 2) {
          v1[i] = v[i + 1];
          v1[i + 1] = -v[i];
        }
      } else {
        for (long long i = 0; i < n; i += 2) {
          v1[i] = v[i + 1];
          v1[i + 1] = -v[i];
        }
      }
      for (auto x : v1) {
        cout << x << " ";
      }
      cout << endl;
    }
  }
  