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;
  void solve(long long nn) {
    long long n;
    cin >> n;
    vector<pair<long long, long long>> vp;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      long long x;
      cin >> x;
      a[i] = x;
      vp.push_back({x, i});
    }
    sort(vp.begin(), vp.end());
    long long second = 0;
    long long iid = -1;
    long long kk = INT_MAX;
    for (long long i = 0; i < n; i++) {
      long long val = vp[i].first;
      if (abs(val) < kk) {
        kk = abs(val);
        iid = i;
      }
    }
    long long k = vp[iid].first;
    long long ans[n];
    for (long long i = n - 1; i >= 0; i--) {
      if (i == iid) continue;
      pair<long long, long long> p = vp[i];
      long long id = p.second;
      long long val = p.first;
      long long zz = abs(val);
      if (i == 1) {
        if (iid == 0) {
          if (second > 0) {
            if (val > 0) {
              ans[id] = kk;
              second += zz * kk;
            } else {
              ans[id] = -kk;
              second += zz * kk;
            }
          } else {
            if (val > 0) {
              ans[id] = -1 * kk;
              second -= zz * kk;
            } else {
              ans[id] = kk;
              second -= zz * kk;
            }
          }
        } else {
          if (second > 0) {
            if (val > 0) {
              ans[id] = -1 * kk;
              second -= zz * kk;
            } else {
              ans[id] = kk;
              second -= zz * kk;
            }
          } else {
            if (val > 0) {
              ans[id] = kk;
              second += zz * kk;
            } else {
              ans[id] = -kk;
              second += zz * kk;
            }
          }
        }
      } else if (i == 0 and iid) {
        if (second > 0) {
          if (val > 0) {
            ans[id] = kk;
            second += zz * kk;
          } else {
            ans[id] = -kk;
            second += zz * kk;
          }
        } else {
          if (val > 0) {
            ans[id] = -1 * kk;
            second -= zz * kk;
          } else {
            ans[id] = kk;
            second -= zz * kk;
          }
        }
      } else {
        if (second > 0) {
          if (val > 0) {
            ans[id] = -1 * kk;
            second -= zz * kk;
          } else {
            ans[id] = kk;
            second -= zz * kk;
          }
        } else {
          if (val > 0) {
            ans[id] = kk;
            second += zz * kk;
          } else {
            ans[id] = -kk;
            second += zz * kk;
          }
        }
      }
    }
    ans[vp[iid].second] = -1 * second / vp[iid].first;
    long long ss = 0;
    for (long long i = 0; i < n; i++) {
      ss += ans[i] * a[i];
      cout << ans[i] << " ";
    }
    cout << "\n";
    return;
  }
  int main() {
    std::cout << std::fixed;
    std::cout << std::setprecision(15);
    ;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ;
    long long t;
    t = 1;
    cin >> t;
    for (long long i = 1; i <= t; i++) solve(i);
    return 0;
  }
  