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.
// Author: Pawan Raj
// Problem: D. Vupsen, Pupsen and 0
// Contest: Codeforces - Codeforces Round #750 (Div. 2)
// URL: https://codeforces.com/contest/1582/problem/D

/* clang-format off */
#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define int long long
#define F(i, n) for (int i = 0; i < n; i++)
#define all(x)(x).begin(), (x).end()
#define show(A) for (auto i: A) cout <<  i << " "; cout << '\n';
#define show_(A, n) F(i, n) { show(A[i]) }
/* clang-format on */

using namespace std;
using ld = long double;
using vi = vector<int>;
using mi = map<int, int>;
using pi = pair<int, int>;

const int N = 300005;
const int MOD = 1e9 + 7;
const int inf = 1e18 + 1;
/***************************code begins here*****************************/

void solve() {
    int n, ans = 0;
    cin >> n;
    vi a(n), b(n);
    F(i, n) cin >> b[i];

    a = b;
    reverse(all(a));

    F(i, n / 2) { a[i] *= -1; }

    if (n % 2) {
        if (abs(a[0]) != abs(b[n / 2])) {
            if (a[0] >= b[n / 2]) {
                a[n / 2] = b[0];
                a[0] -= b[n / 2];
            } else {
                a[n / 2] = -b[0];
                a[0] += b[n / 2];
            }
        } else if (abs(a[n - 1]) != abs(b[n / 2])) {
            if (a[n - 1] >= b[n / 2]) {
                a[n / 2] = b[n - 1];
                a[n - 1] -= b[n / 2];
            } else {
                a[n / 2] = -b[n - 1];
                a[n - 1] += b[n / 2];
            }
        } else {
            if (a[0] >= b[n / 2]) {
                a[n / 2] = b[0];
                a[0] -= b[n / 2];
            } else {
                a[n / 2] = -b[0];
                a[0] += b[n / 2];
            }

            if (a[0] == 0) {
                a[0]--;
                a[n / 2]++;
            }
        }
    }

    int t = 0;
    F(i, n) { t += abs(a[i]); }
    assert(t <= 1e9);

    show(a);
}

main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
}