#include <bits/stdc++.h>
using namespace std;
int a[100010];
int b[100010];
bool IN(int x) { return x >= -10000 && x <= 10000; }
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n;
    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) {
        b[i] = a[i + 1], b[i + 1] = -a[i];
      }
    } else {
      for (int i = 4; i <= n; i += 2) {
        b[i] = a[i + 1], b[i + 1] = -a[i];
      }
      int t1 = a[1] < 0, t2 = a[2] < 0, t3 = a[3] < 0;
      a[1] = abs(a[1]), a[2] = abs(a[2]), a[3] = abs(a[3]);
      if (a[1] == a[2] && a[2] == a[3])
        b[1] = 1, b[2] = 1, b[3] = -2;
      else {
        int flag = 0;
        if (a[1] == a[2]) flag = 1, swap(a[2], a[3]);
        assert(a[1] != a[2]);
        if (IN(a[1] + a[2]))
          b[1] = 1, b[2] = 1;
        else
          assert(IN(a[1] - a[2])), b[1] = 1, b[2] = -1;
        int sum = a[1] * b[1] + a[2] * b[2];
        b[3] = -sum, b[1] *= a[3], b[2] *= a[3];
        if (flag) swap(a[2], a[3]), swap(b[2], b[3]);
      }
      if (t1) b[1] *= -1;
      if (t2) b[2] *= -1;
      if (t3) b[3] *= -1;
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", b[i]);
    }
    printf("\n");
  }
  return 0;
}

### 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;
int a[100010];
int b[100010];
bool IN(int x) { return x >= -10000 && x <= 10000; }
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n;
    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) {
        b[i] = a[i + 1], b[i + 1] = -a[i];
      }
    } else {
      for (int i = 4; i <= n; i += 2) {
        b[i] = a[i + 1], b[i + 1] = -a[i];
      }
      int t1 = a[1] < 0, t2 = a[2] < 0, t3 = a[3] < 0;
      a[1] = abs(a[1]), a[2] = abs(a[2]), a[3] = abs(a[3]);
      if (a[1] == a[2] && a[2] == a[3])
        b[1] = 1, b[2] = 1, b[3] = -2;
      else {
        int flag = 0;
        if (a[1] == a[2]) flag = 1, swap(a[2], a[3]);
        assert(a[1] != a[2]);
        if (IN(a[1] + a[2]))
          b[1] = 1, b[2] = 1;
        else
          assert(IN(a[1] - a[2])), b[1] = 1, b[2] = -1;
        int sum = a[1] * b[1] + a[2] * b[2];
        b[3] = -sum, b[1] *= a[3], b[2] *= a[3];
        if (flag) swap(a[2], a[3]), swap(b[2], b[3]);
      }
      if (t1) b[1] *= -1;
      if (t2) b[2] *= -1;
      if (t3) b[3] *= -1;
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", b[i]);
    }
    printf("\n");
  }
  return 0;
}

```