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;

#define nl '\n'
#define inf LONG_LONG_MAX
#define ninf LONG_LONG_MIN
#define int long long int

const int maxn = 1e5 + 5;
const int modn = 1e9 + 7;
const double pi = acos(-1);

int lcm(int a, int b) {
    return a * b / __gcd(a, b);
}

void solve() {
    int n;
    cin >> n;

    vector<int> a(n);
    for (int i = 0; i < n; i++)
        cin >> a[i];

    vector<int> b(n);
    for (int i = 0; i < n - 3; i += 2) {
        int l = lcm(abs(a[i + 1]), abs(a[i]));
        b[i] = (l / abs(a[i])) * (a[i] < 0 ? -1 : 1);
        b[i + 1] = (l / abs(a[i + 1])) * (a[i + 1] < 0 ? 1 : -1);
    }

    if (n % 2 == 0) {
        int l = lcm(abs(a[n - 2]), abs(a[n - 1]));
        b[n - 2] = (l / abs(a[n - 2])) * (a[n - 2] < 0 ? -1 : 1);
        b[n - 1] = (l / abs(a[n - 1])) * (a[n - 1] < 0 ? 1 : -1);
    } else {
        int l = lcm(abs(a[n - 3]), lcm(abs(a[n - 2]), abs(a[n - 1])));
        b[n - 3] = (l / abs(a[n - 3])) * (a[n - 3] < 0 ? -1 : 1);
        b[n - 2] = (l / abs(a[n - 2])) * (a[n - 2] < 0 ? -1 : 1);
        b[n - 1] = (l / abs(a[n - 1])) * (a[n - 1] < 0 ? 1 : -1);
        b[n - 1] *= 2;
    }

    for (int i : b) {
        cout << i << ' ';
    }

    cout << endl;
}

signed main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int test_case = 1;
    cin >> test_case;
    // compute();
    for (int i = 1; i <= test_case; i++) {
        // cout << "Case #" << i << ": ";
        solve();
    }

    return 0;
}