#include <bits/stdc++.h>
using namespace std;
const long long LLmn = LLONG_MIN;
const long long LLmx = LLONG_MAX;
const long long MOD = 1e6 + 3;
const long long N = 1e9 + 7;
const long double PI = 3.14159265358979323846;
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
pair<int, int> w[9] = {make_pair(-1, 0), make_pair(0, -1),  make_pair(1, 0),
                       make_pair(0, 1),  make_pair(-1, -1), make_pair(0, 1),
                       make_pair(1, -1), make_pair(1, 1)};
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  if (n & 1) {
    if (a[0] + a[1] != 0)
      cout << -a[2] << " " << -a[2] << " " << a[0] + a[1] << " ";
    else if (a[0] + a[2] != 0)
      cout << -a[1] << " " << a[0] + a[2] << " " << -a[1] << " ";
    else
      cout << a[1] + a[2] << " " << -a[0] << " " << -a[0] << " ";
    for (long long i = 3; i < n; i += 2) {
      if (i + 1 < n) cout << a[i + 1] << " " << -a[i] << " ";
    }
    cout << "\n";
    ;
  } else {
    for (long long i = 0; i < n; i += 2) {
      if (i + 1 < n) cout << a[i + 1] << " " << -a[i] << " ";
    }
    cout << "\n";
    ;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your challenge is to write 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;
const long long LLmn = LLONG_MIN;
const long long LLmx = LLONG_MAX;
const long long MOD = 1e6 + 3;
const long long N = 1e9 + 7;
const long double PI = 3.14159265358979323846;
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
pair<int, int> w[9] = {make_pair(-1, 0), make_pair(0, -1),  make_pair(1, 0),
                       make_pair(0, 1),  make_pair(-1, -1), make_pair(0, 1),
                       make_pair(1, -1), make_pair(1, 1)};
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  if (n & 1) {
    if (a[0] + a[1] != 0)
      cout << -a[2] << " " << -a[2] << " " << a[0] + a[1] << " ";
    else if (a[0] + a[2] != 0)
      cout << -a[1] << " " << a[0] + a[2] << " " << -a[1] << " ";
    else
      cout << a[1] + a[2] << " " << -a[0] << " " << -a[0] << " ";
    for (long long i = 3; i < n; i += 2) {
      if (i + 1 < n) cout << a[i + 1] << " " << -a[i] << " ";
    }
    cout << "\n";
    ;
  } else {
    for (long long i = 0; i < n; i += 2) {
      if (i + 1 < n) cout << a[i + 1] << " " << -a[i] << " ";
    }
    cout << "\n";
    ;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) solve();
}

```