#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f;
const double eps = 1e-6;
int a[N];
inline int get(int x, int y) { return x / abs(x) * abs(y); }
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    if (n & 1) {
      for (int i = 1; i <= n - 3; i += 2) {
        cout << get(a[i], a[i + 1]) << ' ' << -get(a[i + 1], a[i]) << ' ';
      }
      cout << get(a[n - 2], a[n]) << ' ' << get(a[n - 1], a[n]) << ' '
           << -get(a[n], ((abs(a[n - 1]) + abs(a[n - 2]))));
    } else {
      for (int i = 1; i <= n; i += 2) {
        cout << get(a[i], a[i + 1]) << ' ' << -get(a[i + 1], a[i]) << ' ';
      }
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f;
const double eps = 1e-6;
int a[N];
inline int get(int x, int y) { return x / abs(x) * abs(y); }
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    if (n & 1) {
      for (int i = 1; i <= n - 3; i += 2) {
        cout << get(a[i], a[i + 1]) << ' ' << -get(a[i + 1], a[i]) << ' ';
      }
      cout << get(a[n - 2], a[n]) << ' ' << get(a[n - 1], a[n]) << ' '
           << -get(a[n], ((abs(a[n - 1]) + abs(a[n - 2]))));
    } else {
      for (int i = 1; i <= n; i += 2) {
        cout << get(a[i], a[i + 1]) << ' ' << -get(a[i + 1], a[i]) << ' ';
      }
    }
    cout << endl;
  }
  return 0;
}

```