#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const int maxn = 1e6 + 10;
inline int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }
int b[maxn];
struct a_node {
  int x;
  int id;
} a[maxn];
bool cmp(a_node x, a_node y) { return abs(x.x) < abs(y.x); }
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].x), b[i] = 0, a[i].id = i;
    sort(a + 1, a + n + 1, cmp);
    if (n % 2) {
      for (int i = 1; i < n - 2; i += 2) {
        int tmp = abs(gcd(a[i].x, a[i + 1].x));
        b[a[i].id] = a[i + 1].x / tmp;
        b[a[i + 1].id] = (-1) * a[i].x / tmp;
      }
      int x = a[n - 1].x + a[n].x, y = a[n - 2].x;
      int tmp = abs(gcd(x, y));
      b[a[n - 2].id] = (-1) * (x) / tmp;
      b[a[n].id] = y / tmp, b[a[n - 1].id] = y / tmp;
      tmp = abs(gcd(a[n - 2].x, a[n].x));
      while (b[a[n - 2].id] == 0) {
        b[a[n - 2].id] -= a[n].x / tmp;
        b[a[n].id] += a[n - 2].x / tmp;
      }
    } else {
      for (int i = 1; i <= n; i += 2) {
        int tmp = abs(gcd(a[i].x, a[i + 1].x));
        b[a[i].id] = a[i + 1].x / tmp;
        b[a[i + 1].id] = -1 * a[i].x / tmp;
      }
    }
    for (int i = 1; i <= n; ++i) {
      printf("%d ", b[i]);
    }
    printf("\n");
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
const double eps = 1e-8;
const int maxn = 1e6 + 10;
inline int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }
int b[maxn];
struct a_node {
  int x;
  int id;
} a[maxn];
bool cmp(a_node x, a_node y) { return abs(x.x) < abs(y.x); }
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].x), b[i] = 0, a[i].id = i;
    sort(a + 1, a + n + 1, cmp);
    if (n % 2) {
      for (int i = 1; i < n - 2; i += 2) {
        int tmp = abs(gcd(a[i].x, a[i + 1].x));
        b[a[i].id] = a[i + 1].x / tmp;
        b[a[i + 1].id] = (-1) * a[i].x / tmp;
      }
      int x = a[n - 1].x + a[n].x, y = a[n - 2].x;
      int tmp = abs(gcd(x, y));
      b[a[n - 2].id] = (-1) * (x) / tmp;
      b[a[n].id] = y / tmp, b[a[n - 1].id] = y / tmp;
      tmp = abs(gcd(a[n - 2].x, a[n].x));
      while (b[a[n - 2].id] == 0) {
        b[a[n - 2].id] -= a[n].x / tmp;
        b[a[n].id] += a[n - 2].x / tmp;
      }
    } else {
      for (int i = 1; i <= n; i += 2) {
        int tmp = abs(gcd(a[i].x, a[i + 1].x));
        b[a[i].id] = a[i + 1].x / tmp;
        b[a[i + 1].id] = -1 * a[i].x / tmp;
      }
    }
    for (int i = 1; i <= n; ++i) {
      printf("%d ", b[i]);
    }
    printf("\n");
  }
  return 0;
}

```