#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using PII = pair<int, int>;
using vi = vector<int>;
const int N = 1e5 + 10;
const ll mod = 1e9 + 7;
int t, n, res[N];
int a[N];
bool st[N];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    if (n % 2 == 0) {
      for (int i = 1; i <= n; i += 2) {
        printf("%d %d ", -a[i + 1], a[i]);
      }
    } else {
      vector<int> v;
      map<int, int> ma;
      for (int i = 4; i <= n; i++) {
        if (ma[a[i]]) {
          res[ma[a[i]]] = -1;
          st[ma[a[i]]] = 1;
          st[i] = 1;
          ma[a[i]] = 0;
          res[i] = 1;
        } else {
          ma[a[i]] = i;
        }
      }
      for (int i = 4; i <= n; i++) {
        if (!st[i]) v.push_back(i);
      }
      int m = v.size();
      for (int i = 0; i < m; i += 2) {
        res[v[i]] = a[v[i + 1]];
        res[v[i + 1]] = -a[v[i]];
      }
      if (a[2] + a[3] != 0) {
        res[1] = a[2] + a[3];
        res[2] = -a[1];
        res[3] = -a[1];
      } else if (a[1] + a[3] != 0) {
        res[2] = a[1] + a[3];
        res[1] = -a[2];
        res[3] = -a[2];
      } else if (a[1] + a[2] != 0) {
        res[3] = a[1] + a[2];
        res[1] = -a[3];
        res[2] = -a[3];
      }
      for (int i = 1; i <= n; i++) printf("%d ", res[i]), st[i] = 0, res[i] = 0;
    }
    puts("");
  }
  return 0;
}

### 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;
using ll = long long;
using PII = pair<int, int>;
using vi = vector<int>;
const int N = 1e5 + 10;
const ll mod = 1e9 + 7;
int t, n, res[N];
int a[N];
bool st[N];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    if (n % 2 == 0) {
      for (int i = 1; i <= n; i += 2) {
        printf("%d %d ", -a[i + 1], a[i]);
      }
    } else {
      vector<int> v;
      map<int, int> ma;
      for (int i = 4; i <= n; i++) {
        if (ma[a[i]]) {
          res[ma[a[i]]] = -1;
          st[ma[a[i]]] = 1;
          st[i] = 1;
          ma[a[i]] = 0;
          res[i] = 1;
        } else {
          ma[a[i]] = i;
        }
      }
      for (int i = 4; i <= n; i++) {
        if (!st[i]) v.push_back(i);
      }
      int m = v.size();
      for (int i = 0; i < m; i += 2) {
        res[v[i]] = a[v[i + 1]];
        res[v[i + 1]] = -a[v[i]];
      }
      if (a[2] + a[3] != 0) {
        res[1] = a[2] + a[3];
        res[2] = -a[1];
        res[3] = -a[1];
      } else if (a[1] + a[3] != 0) {
        res[2] = a[1] + a[3];
        res[1] = -a[2];
        res[3] = -a[2];
      } else if (a[1] + a[2] != 0) {
        res[3] = a[1] + a[2];
        res[1] = -a[3];
        res[2] = -a[3];
      }
      for (int i = 1; i <= n; i++) printf("%d ", res[i]), st[i] = 0, res[i] = 0;
    }
    puts("");
  }
  return 0;
}

```