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.
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<pair<long long, int>> a(n);
    vector<long long> b(n, 1);
    for (int i = 0; i < n; i++) {
      long long x;
      cin >> x;
      a[i] = {x, i};
      if (x < 0) b[i] *= (-1);
    }
    sort(a.begin(), a.end());
    if (n % 2 != 0) {
      pair<long long, int> x0 = a[a.size() - 1];
      pair<long long, int> x1 = a[a.size() - 2];
      pair<long long, int> x2 = a[a.size() - 3];
      long long l = lcm(lcm(abs(x0.first), abs(x1.first)), abs(x2.first));
      b[x0.second] *= (-(l / x0.first) * 2);
      b[x1.second] *= (l / x1.first);
      b[x2.second] *= (l / x2.first);
      n -= 3;
    }
    for (int i = 0; i < n - 1; i += 2) {
      long long l = lcm(abs(a[i].first), abs(a[i + 1].first));
      b[a[i].second] *= (l / a[i].first);
      b[a[i + 1].second] *= (-(l / a[i + 1].first));
    }
    long long sum = 0;
    for (int i = 0; i < (int)b.size(); i++) {
      sum += abs(b[i]);
    }
    assert(sum <= 1e9);
    for (int i = 0; i < (int)b.size(); i++) cout << b[i] << " ";
    cout << "\n";
  }
  return 0;
}
