#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
const int M = N * 2;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const double eps = 1e-7;
const double PI = acos(-1);
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return res;
}
int n, m, k;
int A[N];
char str[N];
signed main() {
  int T = 1;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
    if (n & 1) {
      for (int i = 1; i <= n - 3; i += 2) printf("%d %d ", -A[i + 1], A[i]);
      int a = n, b = n - 1, c = n - 2;
      if (A[b] + A[c] != 0)
        printf("%d %d %d", -A[a], -A[a], A[b] + A[c]);
      else if (A[a] + A[c] != 0)
        printf("%d %d %d", -A[b], A[a] + A[c], -A[b]);
      else if (A[a] + A[b] != 0)
        printf("%d %d %d", A[b] + A[a], -A[c], -A[c]);
    } else {
      for (int i = 1; i <= n; i += 2) printf("%d %d ", -A[i + 1], A[i]);
    }
    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;
const int N = 2e6 + 10;
const int M = N * 2;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const double eps = 1e-7;
const double PI = acos(-1);
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return res;
}
int n, m, k;
int A[N];
char str[N];
signed main() {
  int T = 1;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
    if (n & 1) {
      for (int i = 1; i <= n - 3; i += 2) printf("%d %d ", -A[i + 1], A[i]);
      int a = n, b = n - 1, c = n - 2;
      if (A[b] + A[c] != 0)
        printf("%d %d %d", -A[a], -A[a], A[b] + A[c]);
      else if (A[a] + A[c] != 0)
        printf("%d %d %d", -A[b], A[a] + A[c], -A[b]);
      else if (A[a] + A[b] != 0)
        printf("%d %d %d", A[b] + A[a], -A[c], -A[c]);
    } else {
      for (int i = 1; i <= n; i += 2) printf("%d %d ", -A[i + 1], A[i]);
    }
    puts("");
  }
  return 0;
}

```