#include <bits/stdc++.h>
using namespace std;
namespace {
using T = long long;
namespace module {
inline const T m = 998244353;
void add(T& a, T b) {
  if ((a += b) >= m) a -= m;
}
void dec(T& a, T b) {
  if ((a -= b) < 0) a += m;
}
void mul(T& a, T b) { a = a * b % m; }
T radd(T a, T b) { return (a += b) >= m ? a - m : a; }
T rdec(T a, T b) { return (a -= b) < 0 ? a + m : a; }
T rmul(T a, T b) { return a * b % m; }
};  // namespace module
constexpr char endl = '\n';
const int maxN = 2e5 + 5;
int n;
T a[maxN], b[maxN];
T gcd(T n, T m) { return m ? gcd(m, n % m) : n; }
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i += 2) {
    b[i] = a[i + 1];
    b[i + 1] = -a[i];
    T g = gcd(abs(b[i]), abs(b[i + 1]));
    b[i] /= g;
    b[i + 1] /= g;
  }
  if (n & 1) {
    int i = n - 2, j = n - 1, k = n;
    if (a[j] + a[k] == 0) i = n - 1, j = n - 2, k = n;
    if (a[j] + a[k] == 0) i = n, j = n - 2, k = n - 1;
    b[i] = a[j] + a[k];
    b[j] = -a[i];
    b[k] = -a[i];
    T g = gcd(abs(b[i]), abs(b[j]));
    g = gcd(g, abs(b[k]));
    b[i] /= g;
    b[j] /= g;
    b[k] /= g;
  }
  T sum = 0;
  for (int i = 1; i <= n; i++) sum += a[i] * b[i];
  if (sum) cerr << "NO" << endl;
  for (int i = 1; i <= n; i++) sum += abs(b[i]);
  assert(sum <= 1000000000);
  for (int i = 1; i <= n; i++) cout << b[i] << " \n"[i == n];
}
}  // namespace
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int t(1);
  cin >> t;
  for (int i = 1; i <= t; i++) solve();
}

### Prompt

Create a solution in CPP for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
namespace {
using T = long long;
namespace module {
inline const T m = 998244353;
void add(T& a, T b) {
  if ((a += b) >= m) a -= m;
}
void dec(T& a, T b) {
  if ((a -= b) < 0) a += m;
}
void mul(T& a, T b) { a = a * b % m; }
T radd(T a, T b) { return (a += b) >= m ? a - m : a; }
T rdec(T a, T b) { return (a -= b) < 0 ? a + m : a; }
T rmul(T a, T b) { return a * b % m; }
};  // namespace module
constexpr char endl = '\n';
const int maxN = 2e5 + 5;
int n;
T a[maxN], b[maxN];
T gcd(T n, T m) { return m ? gcd(m, n % m) : n; }
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i += 2) {
    b[i] = a[i + 1];
    b[i + 1] = -a[i];
    T g = gcd(abs(b[i]), abs(b[i + 1]));
    b[i] /= g;
    b[i + 1] /= g;
  }
  if (n & 1) {
    int i = n - 2, j = n - 1, k = n;
    if (a[j] + a[k] == 0) i = n - 1, j = n - 2, k = n;
    if (a[j] + a[k] == 0) i = n, j = n - 2, k = n - 1;
    b[i] = a[j] + a[k];
    b[j] = -a[i];
    b[k] = -a[i];
    T g = gcd(abs(b[i]), abs(b[j]));
    g = gcd(g, abs(b[k]));
    b[i] /= g;
    b[j] /= g;
    b[k] /= g;
  }
  T sum = 0;
  for (int i = 1; i <= n; i++) sum += a[i] * b[i];
  if (sum) cerr << "NO" << endl;
  for (int i = 1; i <= n; i++) sum += abs(b[i]);
  assert(sum <= 1000000000);
  for (int i = 1; i <= n; i++) cout << b[i] << " \n"[i == n];
}
}  // namespace
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int t(1);
  cin >> t;
  for (int i = 1; i <= t; i++) solve();
}

```