#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
void read(vector<int> &a) {
  for (auto &x : a) cin >> x;
}
void read(vector<int64_t> &a) {
  for (auto &x : a) cin >> x;
}
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug &operator<<(const c &) {
    return *this;
  }
};
const int MOD = 1e9 + 7;
const int INF = (int)2e9 + 7;
const int64_t LINF = (int64_t)1e18;
const long double PI = 3.1415926535897932384626433832795;
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  read(v);
  if (n % 2 == 0) {
    for (int i = 0; i < n - 1; i += 2) {
      cout << v[i + 1] << ' ' << -v[i] << ' ';
    }
    cout << "\n";
  } else {
    for (int i = 0; i < n - 3; i += 2) {
      cout << v[i + 1] << ' ' << -v[i] << ' ';
    }
    if (v[n - 1] + v[n - 2] != 0) {
      cout << -(v[n - 1] + v[n - 2]) << ' ' << v[n - 3] << ' ' << v[n - 3]
           << ' ' << "\n";
    } else if (v[n - 1] + v[n - 3] != 0) {
      cout << v[n - 2] << ' ' << -(v[n - 1] + v[n - 3]) << ' ' << v[n - 2]
           << "\n";
    } else {
      cout << v[n - 1] << ' ' << v[n - 1] << ' ' << -(v[n - 3] + v[n - 2])
           << "\n";
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int test_cases = 1;
  cin >> test_cases;
  auto t1 = high_resolution_clock::now();
  for (int tt = 1; tt <= test_cases; tt++) {
    solve();
  }
  auto t2 = high_resolution_clock::now();
  auto time = duration_cast<duration<double>>(t2 - t1);
  cerr << "Time elapsed = " << time.count() << "\n";
}

### 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;
using namespace chrono;
void read(vector<int> &a) {
  for (auto &x : a) cin >> x;
}
void read(vector<int64_t> &a) {
  for (auto &x : a) cin >> x;
}
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug &operator<<(const c &) {
    return *this;
  }
};
const int MOD = 1e9 + 7;
const int INF = (int)2e9 + 7;
const int64_t LINF = (int64_t)1e18;
const long double PI = 3.1415926535897932384626433832795;
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  read(v);
  if (n % 2 == 0) {
    for (int i = 0; i < n - 1; i += 2) {
      cout << v[i + 1] << ' ' << -v[i] << ' ';
    }
    cout << "\n";
  } else {
    for (int i = 0; i < n - 3; i += 2) {
      cout << v[i + 1] << ' ' << -v[i] << ' ';
    }
    if (v[n - 1] + v[n - 2] != 0) {
      cout << -(v[n - 1] + v[n - 2]) << ' ' << v[n - 3] << ' ' << v[n - 3]
           << ' ' << "\n";
    } else if (v[n - 1] + v[n - 3] != 0) {
      cout << v[n - 2] << ' ' << -(v[n - 1] + v[n - 3]) << ' ' << v[n - 2]
           << "\n";
    } else {
      cout << v[n - 1] << ' ' << v[n - 1] << ' ' << -(v[n - 3] + v[n - 2])
           << "\n";
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int test_cases = 1;
  cin >> test_cases;
  auto t1 = high_resolution_clock::now();
  for (int tt = 1; tt <= test_cases; tt++) {
    solve();
  }
  auto t2 = high_resolution_clock::now();
  auto time = duration_cast<duration<double>>(t2 - t1);
  cerr << "Time elapsed = " << time.count() << "\n";
}

```