#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
long long powe(T base, T to) {
  long long ans = 1;
  while (to > 0) {
    if (to & 1)
      ans = ((ans) * (base)), to--;
    else
      base = ((base) * (base)), to /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
bool prime(long long n) {
  if (n == 1) return 0;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
int binarySearch(int arr[], int start, int end, int val) {
  int mid;
  end--;
  while (start <= end) {
    mid = start + (end - start) / 2;
    if (arr[mid] == val)
      return mid;
    else if (arr[mid] < val)
      start = mid + 1;
    else
      end = mid - 1;
  }
  return -1;
}
bool check(string str) {
  bool ans = true;
  for (int i = 0; i < str.size() / 2; i++)
    if (str[i] != str[str.size() - i - 1]) ans = false;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tl = 1;
  cin >> tl;
  while (tl--) {
    int n;
    cin >> n;
    int arr[n];
    for (auto &i : arr) cin >> i;
    ;
    int end = n;
    if (n & 1) end = n - 3;
    vector<int> b(n, 0);
    for (int i = 0; i < end; i += 2) {
      b[i] = -arr[i + 1];
      b[i + 1] = arr[i];
    }
    if (end != n) {
      if ((arr[n - 2] + arr[n - 3]) != 0)
        b[n - 1] = arr[n - 2] + arr[n - 3], b[n - 2] = -arr[n - 1],
              b[n - 3] = -arr[n - 1];
      else if ((arr[n - 2] + arr[n - 1]) != 0)
        b[n - 3] = arr[n - 2] + arr[n - 1], b[n - 2] = -arr[n - 3],
              b[n - 1] = -arr[n - 3];
      else
        b[n - 2] = arr[n - 3] + arr[n - 1], b[n - 3] = -arr[n - 2],
              b[n - 1] = -arr[n - 2];
    }
    for (int i : b) cout << i << " ";
    cout << "\n";
  }
}

### 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>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
long long powe(T base, T to) {
  long long ans = 1;
  while (to > 0) {
    if (to & 1)
      ans = ((ans) * (base)), to--;
    else
      base = ((base) * (base)), to /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
bool prime(long long n) {
  if (n == 1) return 0;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
int binarySearch(int arr[], int start, int end, int val) {
  int mid;
  end--;
  while (start <= end) {
    mid = start + (end - start) / 2;
    if (arr[mid] == val)
      return mid;
    else if (arr[mid] < val)
      start = mid + 1;
    else
      end = mid - 1;
  }
  return -1;
}
bool check(string str) {
  bool ans = true;
  for (int i = 0; i < str.size() / 2; i++)
    if (str[i] != str[str.size() - i - 1]) ans = false;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tl = 1;
  cin >> tl;
  while (tl--) {
    int n;
    cin >> n;
    int arr[n];
    for (auto &i : arr) cin >> i;
    ;
    int end = n;
    if (n & 1) end = n - 3;
    vector<int> b(n, 0);
    for (int i = 0; i < end; i += 2) {
      b[i] = -arr[i + 1];
      b[i + 1] = arr[i];
    }
    if (end != n) {
      if ((arr[n - 2] + arr[n - 3]) != 0)
        b[n - 1] = arr[n - 2] + arr[n - 3], b[n - 2] = -arr[n - 1],
              b[n - 3] = -arr[n - 1];
      else if ((arr[n - 2] + arr[n - 1]) != 0)
        b[n - 3] = arr[n - 2] + arr[n - 1], b[n - 2] = -arr[n - 3],
              b[n - 1] = -arr[n - 3];
      else
        b[n - 2] = arr[n - 3] + arr[n - 1], b[n - 3] = -arr[n - 2],
              b[n - 1] = -arr[n - 2];
    }
    for (int i : b) cout << i << " ";
    cout << "\n";
  }
}

```