#include <bits/stdc++.h>
using namespace std;
double pi = 3.14159265358979323846;
long long mod = 1000000000 + 7;
const int N = 1e5 + 5;
vector<bool> prime(N, true);
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res *= a;
    a *= a;
    b /= 2;
  }
  return res;
}
bool comp(long long a, long long b) { return (a > b); }
void solve() {
  long long n;
  cin >> n;
  vector<long long> vt(n);
  for (long long i = 0; i < n; i += 1) cin >> vt[i];
  vector<long long> b(n);
  for (long long i = 0; i < n - 1; i += 2) {
    if (vt[i] > 0)
      b[i] = abs(vt[i + 1]);
    else
      b[i] = -abs(vt[i + 1]);
    if (vt[i + 1] > 0)
      b[i + 1] = -abs(vt[i]);
    else
      b[i + 1] = abs(vt[i]);
  }
  if (n % 2 == 0) {
    for (auto x : b) cout << x << " ";
  } else {
    long long idx = 0;
    for (long long i = 0; i < n - 1; i += 1) {
      if (vt[i] > 0) {
        if (b[i] + abs(vt[n - 1]) != 0) {
          b[i] += abs(vt[n - 1]);
          idx = i;
          break;
        }
      } else {
        if (b[i] - abs(vt[n - 1]) != 0) {
          b[i] -= abs(vt[n - 1]);
          idx = i;
          break;
        }
      }
    }
    if (vt[n - 1] > 0) {
      b[n - 1] = -abs(vt[idx]);
    } else
      b[n - 1] = abs(vt[idx]);
    for (auto x : b) cout << x << " ";
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please create a solution in cpp 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;
double pi = 3.14159265358979323846;
long long mod = 1000000000 + 7;
const int N = 1e5 + 5;
vector<bool> prime(N, true);
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res *= a;
    a *= a;
    b /= 2;
  }
  return res;
}
bool comp(long long a, long long b) { return (a > b); }
void solve() {
  long long n;
  cin >> n;
  vector<long long> vt(n);
  for (long long i = 0; i < n; i += 1) cin >> vt[i];
  vector<long long> b(n);
  for (long long i = 0; i < n - 1; i += 2) {
    if (vt[i] > 0)
      b[i] = abs(vt[i + 1]);
    else
      b[i] = -abs(vt[i + 1]);
    if (vt[i + 1] > 0)
      b[i + 1] = -abs(vt[i]);
    else
      b[i + 1] = abs(vt[i]);
  }
  if (n % 2 == 0) {
    for (auto x : b) cout << x << " ";
  } else {
    long long idx = 0;
    for (long long i = 0; i < n - 1; i += 1) {
      if (vt[i] > 0) {
        if (b[i] + abs(vt[n - 1]) != 0) {
          b[i] += abs(vt[n - 1]);
          idx = i;
          break;
        }
      } else {
        if (b[i] - abs(vt[n - 1]) != 0) {
          b[i] -= abs(vt[n - 1]);
          idx = i;
          break;
        }
      }
    }
    if (vt[n - 1] > 0) {
      b[n - 1] = -abs(vt[idx]);
    } else
      b[n - 1] = abs(vt[idx]);
    for (auto x : b) cout << x << " ";
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  int t;
  cin >> t;
  while (t--) solve();
}

```