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;
  inline void rd(char *x) { scanf("%s", x); }
  inline void rd(double &x) { scanf("%lf", &x); }
  template <typename T>
  inline void rd(T &x) {
    x = 0;
    bool is = 0;
    char ch = getchar();
    while (!isdigit(ch)) is |= (ch == '-'), ch = getchar();
    while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
    is && (x = -x);
  }
  template <typename T, typename... U>
  inline void rd(T &x, U &...y) {
    rd(x), rd(y...);
  }
  template <typename T>
  inline void write(T x) {
    if (x >= 10) write(x / 10);
    putchar('0' + x % 10);
  }
  const int maxn = 114514;
  long long n, a[maxn], b[maxn];
  long long sgn(long long x) { return x < 0 ? -1 : 1; }
  int main() {
    srand(time(0));
    int T;
    rd(T);
    for (int kase = (1), __kase = (T); kase <= __kase; ++kase) {
      rd(n);
      for (int i = (1), __i = (n); i <= __i; ++i) rd(a[i]);
      if (n % 2 == 0) {
        for (int i = (1), __i = (n / 2); i <= __i; ++i) {
          printf("%lld %lld ", a[i * 2], -a[i * 2 - 1]);
        }
        printf("\n");
        continue;
      }
      long long s2 = 0;
      for (int i = (1), __i = (3); i <= __i; ++i) s2 += a[i];
      b[1] = a[2] + a[3], b[2] = b[3] = -a[1];
      for (int i = (1), __i = (3); i <= __i; ++i) {
        if (s2 - a[i] == 0) continue;
        if ((b[1] == 0 || b[2] == 0 || b[3] == 0) ||
            abs(s2 - 3ll * a[i]) <= abs(b[1] + b[2] + b[3])) {
          b[i] = 0;
          for (int j = (1), __j = (3); j <= __j; ++j)
            if (i != j) b[i] += a[j], b[j] = -a[i];
        }
      }
      if (b[1] == 0 || b[2] == 0 || b[3] == 0)
        while (1)
          ;
      for (int i = (4), __i = (n); i <= __i; ++i) {
        if (i % 2 == 0) {
          b[i] = -a[i + 1], b[i + 1] = a[i];
          if (rand() % 2) b[i] = -b[i], b[i + 1] = -b[i + 1];
        }
      }
      for (int i = (1), __i = (n); i <= __i; ++i) printf("%lld ", b[i]);
      printf("\n");
    }
  }
  