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>
#define pb emplace_back
#define xx first
#define yy second
#define in(x) scanf("%d",&x)
#define lin(x) scanf("%lld",&x)
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int N = 2e6+10;
const int M = N*2;
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const double eps = 1e-7;
const double PI = acos(-1);
ll qpow(ll a,ll b){ll res = 1;while(b){if(b&1)	res = res*a%MOD;a = a*a%MOD;b >>= 1;}return res;}
int n,m,k;
int A[N];
char str[N];
signed main(){
#ifndef ONLINE_JUDGE
    freopen("1.in","r",stdin);
#endif
    int T = 1;  in(T);
    while(T--){
        in(n);
        for(int i = 1;i <= n;i++)   in(A[i]);
        if(n&1){
            for(int i = 1;i <= n-3;i += 2)
                printf("%d %d ",-A[i+1],A[i]);
            int a = n,b = n-1,c = n-2;
            int gcd = __gcd(abs(A[a]),abs(A[b]));
            gcd = __gcd(abs(A[c]),gcd);
            int sum = A[b] + A[c];
            printf("%d %d %d",A[a],A[a],-sum);
        }else{
            for(int i = 1;i <= n;i += 2)
                printf("%d %d ",-A[i+1],A[i]);
        }
        puts("");
    }
    return 0;
}