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 int MAX = 1e5 + 5;
  long long t, n, a[MAX];
  long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
  int main() {
    long long i, j, f1, f2, g1, g2, gg;
    scanf("%lld", &t);
    while (t--) {
      scanf("%lld", &n);
      for (i = 1; i <= n; i++) scanf("%lld", a + i);
      if (n % 2 == 0) {
        for (i = 1; i <= n; i += 2) {
          f1 = f2 = 0;
          g1 = g2 = 1;
          if (a[i] < 0) a[i] = -a[i], f1 = 1;
          if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
          gg = gcd(a[i], a[i + 1]);
          if (f1 ^ f2)
            printf("%lld %lld ", a[i + 1] / gg, a[i] / gg);
          else
            printf("%lld %lld ", -a[i + 1] / gg, a[i] / gg);
        }
        printf("\n");
      } else {
        for (i = 1; i <= n - 3; i += 2) {
          f1 = f2 = 0;
          g1 = g2 = 1;
          if (a[i] < 0) a[i] = -a[i], f1 = 1;
          if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
          gg = gcd(a[i], a[i + 1]);
          if (f1 ^ f2)
            printf("%lld %lld ", a[i + 1] / gg, a[i] / gg);
          else
            printf("%lld %lld ", -a[i + 1] / gg, a[i] / gg);
        }
        if (a[i] == a[i + 1]) {
          f1 = f2 = 0;
          i++;
          if (a[i] < 0) a[i] = -a[i], f1 = 1;
          if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
          gg = gcd(a[i], a[i + 1]);
          if (f1 ^ f2)
            printf("-1 %lld %lld ", a[i + 1] / gg + 1, a[i] / gg);
          else
            printf("1 %lld %lld ", -(a[i + 1] / gg + 1), a[i] / gg);
          printf("\n");
          continue;
        }
        if (a[i] == -a[i + 1]) {
          f1 = f2 = 0;
          i++;
          if (a[i] < 0) a[i] = -a[i], f1 = 1;
          if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
          gg = gcd(a[i], a[i + 1]);
          if (f1 ^ f2)
            printf("1 %lld %lld ", a[i + 1] / gg + 1, a[i] / gg);
          else
            printf("-1 %lld %lld ", -(a[i + 1] / gg + 1), a[i] / gg);
          printf("\n");
          continue;
        }
        long long kk = 1;
        if ((a[i + 1] < 0 && a[i] > 0) || (a[i + 1] > 0 && a[i] < 0)) {
          a[i + 1] += a[i];
          i++;
        } else {
          a[i + 1] -= a[i];
          i++;
          kk = -1;
        }
        f1 = f2 = 0;
        if (a[i] < 0) a[i] = -a[i], f1 = 1;
        if (a[i + 1] < 0) a[i + 1] = -a[i + 1], f2 = 1;
        gg = gcd(a[i], a[i + 1]);
        if (f1 ^ f2)
          printf("%lld %lld %lld ", a[i + 1] * kk / gg, a[i + 1] / gg, a[i] / gg);
        else
          printf("%lld %lld %lld ", -a[i + 1] * kk / gg, -a[i + 1] / gg,
                 a[i] / gg);
        printf("\n");
      }
    }
    return 0;
  }
  