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 t, n, a[100005], b[100005], wz[100005];
  const long long INF = 1e9;
  long long sum;
  int jdz(int x) {
    if (x < 0) return -x;
    return x;
  }
  bool cmp(int x, int y) {
    if (jdz(a[x]) == jdz(a[y])) return a[x] < a[y];
    return jdz(a[x]) < jdz(a[y]);
  }
  int main() {
    scanf("%d", &t);
    while (t--) {
      scanf("%d", &n);
      for (int i = 1; i <= n; i++) {
        wz[i] = i;
        scanf("%d", &a[i]);
      }
      sort(wz + 1, wz + n + 1, cmp);
      if (n & 1) {
        if ((a[wz[2]] + a[wz[3]]) == 0) swap(wz[1], wz[3]);
        b[wz[1]] = a[wz[2]] + a[wz[3]];
        b[wz[2]] = -a[wz[1]];
        b[wz[3]] = -a[wz[1]];
        for (int i = 4; i <= n; i += 2) {
          if (jdz(a[wz[i]]) == jdz(a[wz[i + 1]])) {
            if (a[wz[i]] == a[wz[i + 1]]) {
              b[wz[i]] = 1;
              b[wz[i + 1]] = -1;
            } else {
              b[wz[i]] = 1;
              b[wz[i + 1]] = 1;
            }
          } else {
            b[wz[i]] = -a[wz[i + 1]];
            b[wz[i + 1]] = a[wz[i]];
          }
        }
        sum = 0;
        for (int i = 1; i <= n; i++) {
          sum = sum + jdz(b[i]);
        }
        if (sum > INF) {
          for (int i = 5; i < n; i += 2) {
            if (jdz(a[wz[i]]) == jdz(a[wz[i + 1]])) {
              if (a[wz[i]] == a[wz[i + 1]]) {
                b[wz[i]] = 1;
                b[wz[i + 1]] = -1;
              } else {
                b[wz[i]] = 1;
                b[wz[i + 1]] = 1;
              }
            } else {
              b[wz[i]] = -a[wz[i + 1]];
              b[wz[i + 1]] = a[wz[i]];
            }
          }
          if (jdz(a[wz[4]]) == jdz(a[wz[n]])) {
            if (a[wz[4]] == a[wz[n]]) {
              b[wz[4]] = 1;
              b[wz[n]] = -1;
            } else {
              b[wz[4]] = 1;
              b[wz[n]] = 1;
            }
          } else {
            b[wz[4]] = -a[wz[n]];
            b[wz[n]] = a[wz[4]];
          }
        }
        for (int i = 1; i <= n; i++) printf("%d ", b[i]);
      } else {
        for (int i = 1; i <= n; i += 2) {
          printf("%d %d ", -a[i + 1], a[i]);
        }
      }
      printf("\n");
    }
    return 0;
  }
  