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 = 1e5+10;
const int M = 1e4+10;
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],ans[N];
int st[M];
signed main(){
#ifndef ONLINE_JUDGE
    freopen("1.in","r",stdin);
#endif
    int T = 1;  in(T);
    while(T--){
        in(n);
        memset(st,0,sizeof st);
        memset(ans,0,sizeof ans);
        for(int i = 1;i <= n;i++){
            ans[i] = 0;
            in(A[i]);
            int tp = abs(A[i]);
            if(!st[tp]){
                st[tp] = (A[i] < 0 ? -1 : 1) * i;
            }else{  // 如果前面出现过 不论正负
                if(A[i]*st[tp] > 0){    // 两个同号 需要一正一负
                    ans[abs(st[tp])] = 1,ans[i] = -1;
                }else{  // 异号 全正
                    ans[abs(st[tp])] = 1,ans[i] = 1;
                }
                st[tp] = 0;
            }
        }
        int sum = 0,gcd = 0;
        for(int i = 1;i <= n;i++){
            if(!ans[i]) sum += A[i],gcd = __gcd(gcd,abs(A[i]));
        }
        if(!sum){
            for(int i = 1;i <= n;i++)
                printf("%d ",ans[i]);
            puts("");
            continue;
        }
        sum /= gcd;
        int mn = INF;
        for(int i = 1;i <= n;i++)
            if(!ans[i]) mn = min(abs(A[i])/gcd,mn);

        ll check = 0;
        bool ok = 0;
        for(int i = 1;i <= n;i++){
            if(ans[i]) continue;
            if(!ok && abs(A[i]) == mn*gcd){
                int t;
                if(A[i] < 0)    t = (sum + mn);
                else            t = -(sum - mn);
                ok = 1;
                check += t*A[i];
                printf("%d ",t);
            }else{
                printf("%d ",mn);
                check += mn*A[i];
            }
        }
        puts("");
        //cout << check << endl;
    }
    return 0;
}
