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;
using ll  =  long long;
using ld  =  long double;
const ll   MOD        =  1e+9 + 7;

void MAIN() {
    int n; cin >> n;
    vector<ll> v(n);
    for (auto &x : v) cin >> x;
    ll lcm;
    vector<int> b(n);
    for (int i = 0; i < n - 3; i += 2) {
        lcm = abs(v[i]) * abs(v[i + 1]) / __gcd(abs(v[i]), abs(v[i + 1]));
        b[i] = (lcm / abs(v[i])) * (v[i] < 0 ? -1 : 1);
        b[i + 1] = (lcm / abs(v[i + 1])) * (v[i + 1] < 0 ? 1 : -1);
    }
    if (n % 2) {
        lcm = abs(v[n - 1]) * abs(v[n - 2]) / __gcd(abs(v[n - 1]), abs(v[n - 2]));
        lcm = abs(v[n - 3]) * lcm / __gcd(lcm, abs(v[n - 3]));
        b[n - 3] = (lcm / abs(v[n - 3])) * (v[n - 3] < 0 ? 1 : -1);
        b[n - 2] = (lcm / abs(v[n - 2])) * (v[n - 2] < 0 ? 1 : -1);
        b[n - 1] = 2 * (lcm / abs(v[n - 1])) * (v[n - 1] < 0 ? -1 : 1);
    }
    else {
        int i = n - 2;
        lcm = abs(v[i]) * abs(v[i + 1]) / __gcd(abs(v[i]), abs(v[i + 1]));
        b[i] = (lcm / abs(v[i])) * (v[i] < 0 ? -1 : 1);
        b[i + 1] = (lcm / abs(v[i + 1])) * (v[i + 1] < 0 ? 1 : -1);
    }

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

    cout << '\n';
}

int main() {
    ios:: sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
    cout << fixed; cout << setprecision(10);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    int t;
    cin >> t;
    while (t--) MAIN();
}