#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
  ll n;
  cin >> n;
  vector<ll> v(n);
  vector<array<ll, 2>> t;
  vector<ll> res(n);
  for (ll i = 0; i < n; i++) {
    cin >> v[i];
    t.push_back({v[i], i});
  }
  sort(t.begin(), t.end());
  while (!t.empty()) {
    ll m = t.size();
    if (m == 3) {
      array<ll, 2> a = t.back();
      t.pop_back();
      array<ll, 2> b = t.back();
      t.pop_back();
      array<ll, 2> c = t.back();
      t.pop_back();
      if (b[0] == c[0]) swap(a, c);
      if (a[0] == b[0]) {
        res[a[1]] = c[0];
        res[b[1]] = -2 * c[0];
        res[c[1]] = b[0];
      } else {
        res[a[1]] = b[0] - c[0];
        res[b[1]] = c[0] - a[0];
        res[c[1]] = a[0] - b[0];
      }
      break;
    }
    array<ll, 2> p1 = t.back();
    t.pop_back();
    array<ll, 2> p2 = t.back();
    t.pop_back();
    if (p1[0] == p2[0]) {
      res[p1[1]] = -1;
      res[p2[1]] = 1;
    } else {
      res[p1[1]] = -p2[0];
      res[p2[1]] = p1[0];
    }
  }
  for (auto x : res) cout << x << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int _t = 1;
  cin >> _t;
  for (int i = 1; i <= _t; i++) {
    solve();
  }
  {};
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
using ll = long long;
void solve() {
  ll n;
  cin >> n;
  vector<ll> v(n);
  vector<array<ll, 2>> t;
  vector<ll> res(n);
  for (ll i = 0; i < n; i++) {
    cin >> v[i];
    t.push_back({v[i], i});
  }
  sort(t.begin(), t.end());
  while (!t.empty()) {
    ll m = t.size();
    if (m == 3) {
      array<ll, 2> a = t.back();
      t.pop_back();
      array<ll, 2> b = t.back();
      t.pop_back();
      array<ll, 2> c = t.back();
      t.pop_back();
      if (b[0] == c[0]) swap(a, c);
      if (a[0] == b[0]) {
        res[a[1]] = c[0];
        res[b[1]] = -2 * c[0];
        res[c[1]] = b[0];
      } else {
        res[a[1]] = b[0] - c[0];
        res[b[1]] = c[0] - a[0];
        res[c[1]] = a[0] - b[0];
      }
      break;
    }
    array<ll, 2> p1 = t.back();
    t.pop_back();
    array<ll, 2> p2 = t.back();
    t.pop_back();
    if (p1[0] == p2[0]) {
      res[p1[1]] = -1;
      res[p2[1]] = 1;
    } else {
      res[p1[1]] = -p2[0];
      res[p2[1]] = p1[0];
    }
  }
  for (auto x : res) cout << x << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int _t = 1;
  cin >> _t;
  for (int i = 1; i <= _t; i++) {
    solve();
  }
  {};
  return 0;
}

```