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>
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long llINF = 9223372036854775807;
const int N = 1e5 + 10;
const double pi = acos(-1.0);
using namespace std;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long qpow(long long b, long long e, long long m = mod) {
  long long a = 1;
  for (; e; e >>= 1, b = b * b % m)
    if (e & 1) a = a * b % m;
  return a;
}
long long n;
long long a[N], res[N];
void solve() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  long long g;
  if (n % 2 == 0) {
    for (int i = 1; i <= n; i += 2) {
      res[i] = -a[i + 1];
      res[i + 1] = a[i];
    }
  } else {
    for (int i = 1; i <= n - 3; i += 2) {
      res[i] = -a[i + 1];
      res[i + 1] = a[i];
    }
    long long z = lcm(a[n - 2], a[n - 1]);
    z = lcm(z, a[n]);
    res[n - 2] = z / a[n - 2], res[n - 1] = z / a[n - 1],
            res[n] = -2 * z / a[n];
  }
  for (int i = 1; i <= n; i++) {
    if (i < n)
      printf("%lld ", res[i]);
    else
      printf("%lld\n", res[i]);
  }
}
int main() {
  int _ = 1;
  scanf("%d", &_);
  while (_--) {
    solve();
  }
  return 0;
}
