#include <bits/stdc++.h>
using namespace std;
int f(int a, long long int* d) {
  if (a == d[a]) return a;
  return d[a] = f(d[a], d);
}
void u(int a, int b, long long int* d, long long int* s) {
  a = f(a, d);
  b = f(b, d);
  if (a != b) {
    d[b] = a;
    s[a] += s[b];
  }
}
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long int binpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int main() {
  int w;
  long long int a, b, n, c;
  long long int t;
  cin >> w;
  while (w--) {
    cin >> n;
    int d[n + 1];
    for (int i = 0; i < n; i++) cin >> d[i];
    int p[n + 1];
    if (n % 2 == 0) {
      for (int i = 0; i < n; i++) {
        if (i % 2 == 0)
          p[i] = -d[i + 1];
        else
          p[i] = d[i - 1];
      }
    } else {
      for (int i = 0; i < n - 3; i++) {
        if (i % 2 == 0)
          p[i] = -d[i + 1];
        else
          p[i] = d[i - 1];
      }
      int f = 0;
      for (int i = n - 3; i <= n - 1; i++) {
        for (int j = n - 3; j <= n - 1; j++) {
          if (i != j && (d[i] + d[j]) != 0) {
            int k;
            for (k = n - 3; k <= n - 1; k++) {
              if (k != i && k != j) break;
            }
            p[k] = -(d[i] + d[j]);
            p[i] = d[k];
            p[j] = d[k];
            f = 1;
            break;
          }
        }
        if (f) break;
      }
    }
    for (int i = 0; i < n; i++) cout << p[i] << " ";
    cout << "\n";
  }
}

### Prompt

Your task is to create 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;
int f(int a, long long int* d) {
  if (a == d[a]) return a;
  return d[a] = f(d[a], d);
}
void u(int a, int b, long long int* d, long long int* s) {
  a = f(a, d);
  b = f(b, d);
  if (a != b) {
    d[b] = a;
    s[a] += s[b];
  }
}
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long int binpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int main() {
  int w;
  long long int a, b, n, c;
  long long int t;
  cin >> w;
  while (w--) {
    cin >> n;
    int d[n + 1];
    for (int i = 0; i < n; i++) cin >> d[i];
    int p[n + 1];
    if (n % 2 == 0) {
      for (int i = 0; i < n; i++) {
        if (i % 2 == 0)
          p[i] = -d[i + 1];
        else
          p[i] = d[i - 1];
      }
    } else {
      for (int i = 0; i < n - 3; i++) {
        if (i % 2 == 0)
          p[i] = -d[i + 1];
        else
          p[i] = d[i - 1];
      }
      int f = 0;
      for (int i = n - 3; i <= n - 1; i++) {
        for (int j = n - 3; j <= n - 1; j++) {
          if (i != j && (d[i] + d[j]) != 0) {
            int k;
            for (k = n - 3; k <= n - 1; k++) {
              if (k != i && k != j) break;
            }
            p[k] = -(d[i] + d[j]);
            p[i] = d[k];
            p[j] = d[k];
            f = 1;
            break;
          }
        }
        if (f) break;
      }
    }
    for (int i = 0; i < n; i++) cout << p[i] << " ";
    cout << "\n";
  }
}

```