#include <bits/stdc++.h>
using namespace std;
int gcdeuclid(int a, int b) {
  if (a == 0) {
    return b;
  }
  return gcdeuclid(b % a, a);
}
void solve() {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int ans[n];
  if (n % 2 == 0) {
    int count = 0;
    int i = 0;
    int j = n - 1;
    while (count <= n / 2) {
      ans[i] = -a[j];
      ans[j] = a[i];
      i++;
      j--;
      count++;
    }
  } else {
    int count = 0;
    int i = 0;
    int j = n - 1;
    while (count < ((n / 2) - 1)) {
      ans[i] = -a[j];
      ans[j] = a[i];
      i++;
      j--;
      count++;
    }
    if (a[i] + a[i + 1] != 0) {
      ans[i] = a[j];
      ans[i + 1] = a[j];
      ans[j] = -(a[i] + a[i + 1]);
    } else {
      if (a[i] - a[j] != 0) {
        ans[i] = -a[i + 1];
        ans[j] = a[i + 1];
        ans[i + 1] = (a[i] - a[j]);
      } else {
        ans[i] = a[i + 1];
        ans[j] = a[i + 1];
        ans[i + 1] = -(a[i] + a[j]);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    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;
int gcdeuclid(int a, int b) {
  if (a == 0) {
    return b;
  }
  return gcdeuclid(b % a, a);
}
void solve() {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int ans[n];
  if (n % 2 == 0) {
    int count = 0;
    int i = 0;
    int j = n - 1;
    while (count <= n / 2) {
      ans[i] = -a[j];
      ans[j] = a[i];
      i++;
      j--;
      count++;
    }
  } else {
    int count = 0;
    int i = 0;
    int j = n - 1;
    while (count < ((n / 2) - 1)) {
      ans[i] = -a[j];
      ans[j] = a[i];
      i++;
      j--;
      count++;
    }
    if (a[i] + a[i + 1] != 0) {
      ans[i] = a[j];
      ans[i + 1] = a[j];
      ans[j] = -(a[i] + a[i + 1]);
    } else {
      if (a[i] - a[j] != 0) {
        ans[i] = -a[i + 1];
        ans[j] = a[i + 1];
        ans[i + 1] = (a[i] - a[j]);
      } else {
        ans[i] = a[i + 1];
        ans[j] = a[i + 1];
        ans[i + 1] = -(a[i] + a[j]);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```