#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> solve(int x, int y, int z) {
  vector<pair<int, int>> v = vector<pair<int, int>>{pair<int, int>(abs(x), x),
                                                    pair<int, int>(abs(y), y),
                                                    pair<int, int>(abs(z), z)};
  sort(v.begin(), v.end());
  if (v[2].first % v[0].first != 0) {
    int d = v[2].first / v[0].first, r = v[2].first % v[0].first;
    vector<pair<int, int>> m = vector<pair<int, int>>();
    m.push_back(
        pair<int, int>(v[0].second, d * v[1].first * v[0].second / v[0].first));
    m.push_back(pair<int, int>(v[1].second, r * v[1].second / v[1].first));
    m.push_back(
        pair<int, int>(v[2].second, -v[1].first * v[2].second / v[2].first));
    return m;
  }
  if (v[2].first % v[1].first != 0) {
    int d = v[2].first / v[1].first, r = v[2].first % v[1].first;
    vector<pair<int, int>> m = vector<pair<int, int>>();
    m.push_back(
        pair<int, int>(v[1].second, d * v[0].first * v[1].second / v[1].first));
    m.push_back(pair<int, int>(v[0].second, r * v[0].second / v[0].first));
    m.push_back(
        pair<int, int>(v[2].second, -v[0].first * v[2].second / v[2].first));
    return m;
  } else {
    vector<pair<int, int>> m = vector<pair<int, int>>();
    m.push_back(pair<int, int>(v[2].second, -2));
    m.push_back(pair<int, int>(v[1].second, v[2].second / v[1].second));
    m.push_back(pair<int, int>(v[0].second, v[2].second / v[0].second));
    return m;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t, n, a, b, c;
  cin >> t;
  while (t--) {
    cin >> n;
    if (n % 2) {
      n -= 3;
      cin >> a >> b >> c;
      vector<pair<int, int>> m = solve(a, b, c);
      set<int> s = set<int>{0, 1, 2};
      for (int x : {a, b, c}) {
        for (int i : s) {
          if (m[i].first == x) {
            cout << m[i].second << ' ';
            s.erase(i);
            break;
          }
        }
      }
    }
    while (n) {
      n -= 2;
      cin >> a >> b;
      cout << -b << ' ' << a << ' ';
    }
    cout << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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;
vector<pair<int, int>> solve(int x, int y, int z) {
  vector<pair<int, int>> v = vector<pair<int, int>>{pair<int, int>(abs(x), x),
                                                    pair<int, int>(abs(y), y),
                                                    pair<int, int>(abs(z), z)};
  sort(v.begin(), v.end());
  if (v[2].first % v[0].first != 0) {
    int d = v[2].first / v[0].first, r = v[2].first % v[0].first;
    vector<pair<int, int>> m = vector<pair<int, int>>();
    m.push_back(
        pair<int, int>(v[0].second, d * v[1].first * v[0].second / v[0].first));
    m.push_back(pair<int, int>(v[1].second, r * v[1].second / v[1].first));
    m.push_back(
        pair<int, int>(v[2].second, -v[1].first * v[2].second / v[2].first));
    return m;
  }
  if (v[2].first % v[1].first != 0) {
    int d = v[2].first / v[1].first, r = v[2].first % v[1].first;
    vector<pair<int, int>> m = vector<pair<int, int>>();
    m.push_back(
        pair<int, int>(v[1].second, d * v[0].first * v[1].second / v[1].first));
    m.push_back(pair<int, int>(v[0].second, r * v[0].second / v[0].first));
    m.push_back(
        pair<int, int>(v[2].second, -v[0].first * v[2].second / v[2].first));
    return m;
  } else {
    vector<pair<int, int>> m = vector<pair<int, int>>();
    m.push_back(pair<int, int>(v[2].second, -2));
    m.push_back(pair<int, int>(v[1].second, v[2].second / v[1].second));
    m.push_back(pair<int, int>(v[0].second, v[2].second / v[0].second));
    return m;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t, n, a, b, c;
  cin >> t;
  while (t--) {
    cin >> n;
    if (n % 2) {
      n -= 3;
      cin >> a >> b >> c;
      vector<pair<int, int>> m = solve(a, b, c);
      set<int> s = set<int>{0, 1, 2};
      for (int x : {a, b, c}) {
        for (int i : s) {
          if (m[i].first == x) {
            cout << m[i].second << ' ';
            s.erase(i);
            break;
          }
        }
      }
    }
    while (n) {
      n -= 2;
      cin >> a >> b;
      cout << -b << ' ' << a << ' ';
    }
    cout << '\n';
  }
  return 0;
}

```