#include <bits/stdc++.h>
namespace FastRead {
char __buff[5000];
int __lg = 0, __p = 0;
char nc() {
  if (__lg == __p) {
    __lg = fread(__buff, 1, 5000, stdin);
    __p = 0;
    if (!__lg) return EOF;
  }
  return __buff[__p++];
}
template <class T>
void read(T& __x) {
  T __sgn = 1;
  char __c;
  while (!isdigit(__c = nc()))
    if (__c == '-') __sgn = -1;
  __x = __c - '0';
  while (isdigit(__c = nc())) __x = __x * 10 + __c - '0';
  __x *= __sgn;
}
}  // namespace FastRead
using namespace FastRead;
using namespace std;
const int N = 1e5 + 10;
const int M = 1e9 + 7;
const long double PI = acos(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n;
int a[N], b[N];
void Test() {
  int i;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i += 2) {
    b[i] = a[i + 1];
    b[i + 1] = a[i];
    b[i + 1] *= -1;
  }
  if (n & 1) {
    b[n - 1] = b[n - 2] = a[n];
    b[n] = a[n - 1] + a[n - 2];
    if (!b[n]) {
      b[n - 1] *= 2;
      b[n] = 2 * a[n - 1] + a[n - 2];
    }
    b[n] *= -1;
  }
  for (i = 1; i <= n; i++) cout << b[i] << ' ';
  cout << '\n';
}
signed main() {
  int q;
  cin >> q;
  while (q--) Test();
  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>
namespace FastRead {
char __buff[5000];
int __lg = 0, __p = 0;
char nc() {
  if (__lg == __p) {
    __lg = fread(__buff, 1, 5000, stdin);
    __p = 0;
    if (!__lg) return EOF;
  }
  return __buff[__p++];
}
template <class T>
void read(T& __x) {
  T __sgn = 1;
  char __c;
  while (!isdigit(__c = nc()))
    if (__c == '-') __sgn = -1;
  __x = __c - '0';
  while (isdigit(__c = nc())) __x = __x * 10 + __c - '0';
  __x *= __sgn;
}
}  // namespace FastRead
using namespace FastRead;
using namespace std;
const int N = 1e5 + 10;
const int M = 1e9 + 7;
const long double PI = acos(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n;
int a[N], b[N];
void Test() {
  int i;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i += 2) {
    b[i] = a[i + 1];
    b[i + 1] = a[i];
    b[i + 1] *= -1;
  }
  if (n & 1) {
    b[n - 1] = b[n - 2] = a[n];
    b[n] = a[n - 1] + a[n - 2];
    if (!b[n]) {
      b[n - 1] *= 2;
      b[n] = 2 * a[n - 1] + a[n - 2];
    }
    b[n] *= -1;
  }
  for (i = 1; i <= n; i++) cout << b[i] << ' ';
  cout << '\n';
}
signed main() {
  int q;
  cin >> q;
  while (q--) Test();
  return 0;
}

```