#include <bits/stdc++.h>
using namespace std;
int main() {
  int t, sum, n, kl1 = 0, kl0 = 0;
  cin >> t;
  while (t--) {
    cin >> n;
    int* a = new int[n];
    int* b = new int[n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      b[i] = 1;
    }
    if (n % 2 == 0) {
      for (int i = 0; i < n - 1; i = i + 2) {
        if (a[i] * a[i + 1] < 0) {
          if (a[i] < 0) {
            b[i] = a[i + 1];
            b[i + 1] = -1 * a[i];
          } else {
            b[i] = -1 * a[i + 1];
            b[i + 1] = a[i];
          }
        } else {
          b[i] = -1 * a[i + 1];
          b[i + 1] = a[i];
        }
      }
    } else {
      if (n > 3) {
        for (int i = 0; i < n - 4; i = i + 2) {
          if (a[i] * a[i + 1] < 0) {
            if (a[i] < 0) {
              b[i] = a[i + 1];
              b[i + 1] = -1 * a[i];
            } else {
              b[i] = -1 * a[i + 1];
              b[i + 1] = a[i];
            }
          } else {
            b[i] = -1 * a[i + 1];
            b[i + 1] = a[i];
          }
        }
      }
      if (a[n - 3] < 0) b[n - 3] *= -1;
      if (a[n - 2] < 0) b[n - 2] *= -1;
      b[n - 3] *= abs(a[n - 1]);
      b[n - 2] *= abs(a[n - 1]);
      if (a[n - 1] < 0) {
        b[n - 1] = abs(a[n - 3]) + abs(a[n - 2]);
      } else {
        b[n - 1] = (abs(a[n - 3]) + abs(a[n - 2])) * (-1);
      }
    }
    for (int i = 0; i < n; i++) {
      cout << b[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}

### Prompt

Generate a CPP solution to 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 main() {
  int t, sum, n, kl1 = 0, kl0 = 0;
  cin >> t;
  while (t--) {
    cin >> n;
    int* a = new int[n];
    int* b = new int[n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      b[i] = 1;
    }
    if (n % 2 == 0) {
      for (int i = 0; i < n - 1; i = i + 2) {
        if (a[i] * a[i + 1] < 0) {
          if (a[i] < 0) {
            b[i] = a[i + 1];
            b[i + 1] = -1 * a[i];
          } else {
            b[i] = -1 * a[i + 1];
            b[i + 1] = a[i];
          }
        } else {
          b[i] = -1 * a[i + 1];
          b[i + 1] = a[i];
        }
      }
    } else {
      if (n > 3) {
        for (int i = 0; i < n - 4; i = i + 2) {
          if (a[i] * a[i + 1] < 0) {
            if (a[i] < 0) {
              b[i] = a[i + 1];
              b[i + 1] = -1 * a[i];
            } else {
              b[i] = -1 * a[i + 1];
              b[i + 1] = a[i];
            }
          } else {
            b[i] = -1 * a[i + 1];
            b[i + 1] = a[i];
          }
        }
      }
      if (a[n - 3] < 0) b[n - 3] *= -1;
      if (a[n - 2] < 0) b[n - 2] *= -1;
      b[n - 3] *= abs(a[n - 1]);
      b[n - 2] *= abs(a[n - 1]);
      if (a[n - 1] < 0) {
        b[n - 1] = abs(a[n - 3]) + abs(a[n - 2]);
      } else {
        b[n - 1] = (abs(a[n - 3]) + abs(a[n - 2])) * (-1);
      }
    }
    for (int i = 0; i < n; i++) {
      cout << b[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}

```