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;
const int N=1e5+10;
int b[N];
typedef long long ll;
struct stu
{
	int val,id;
	bool operator<(const stu &W)const
	{
		return abs(val)<abs(W.val);
	}
} a[N];
ll exgcd(ll a,ll b,ll &x,ll &y)
{
	if(b == 0)
	{
		x = 1;
		y = 0;
		return a;
	}
	ll d = exgcd(b,a%b,x,y);  //d 就是gcd(a,b)
	ll t = x;
	x = y;
	y = t-(a/b)*y;
	return d;
}
bool linear_equation(ll a,ll b,ll c,ll &x,ll &y)
{
	ll d = exgcd(a,b,x,y);  //d是gcd(a,b)//求出a*x+b*y = gcd(a,b)的一组解

	if(c%d) return false;
	ll k = c/d;    //上述解乘以 c/gcd(a,b)就是 a*x +b*y = c的解
	x *= k;
	y *= k;    //求得的只是其中一组解
	return true;
}
int gcd(int x,int y)
{
	x=abs(x),y=abs(y);
//	printf("%d      %d  %d\n",x,y,__gcd(x,y));
	return __gcd(x,y);
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		for(int i=1; i<=n; i++)
		{
			scanf("%d",&a[i].val);
			a[i].id=i;
		}
		sort(a+1,a+1+n);
		if(n%2==0)
		{
			for(int i=1; i<=n; i+=2)
			{
				if((abs(a[i].val)==a[i].val&&abs(a[i+1].val)!=a[i+1].val)||abs(a[i].val)!=a[i].val&&abs(a[i+1].val)==a[i+1].val)
				{
					int x=a[i+1].val,y=a[i].val;
					b[a[i].id]=x/gcd(x,y),b[a[i-1].id]=y/gcd(x,y);
				}
				else
				{
					int x=a[i+1].val,y=a[i].val;
					b[a[i].id]=-x/gcd(x,y),b[a[i+1].id]=y/gcd(x,y);
				}
			}
		}
		else
		{
			for(int i=n; i>3; i-=2)
			{
				if((abs(a[i].val)==a[i].val&&abs(a[i-1].val)!=a[i-1].val)||abs(a[i].val)!=a[i].val&&abs(a[i-1].val)==a[i-1].val)
				{
					int x=a[i-1].val,y=a[i].val;
//					printf("%d  %d       %d\n",x,y,gcd(x,y));
					b[a[i].id]=x/gcd(x,y),b[a[i-1].id]=y/gcd(x,y);
				}
				else
				{
					int x=a[i-1].val,y=a[i].val;
					b[a[i].id]=-x/gcd(x,y),b[a[i-1].id]=y/gcd(x,y);
				}
			}
			ll x=a[1].val,y=a[2].val,z=a[3].val;
			for(int i=1; i<=1000; i++)
			{
				ll c,d;
				if(!linear_equation(x,y,z*i,c,d))continue;
				else
				{
					b[a[1].id]=c,b[a[2].id]=d,b[a[3].id]=-i;
					break;
				}
			}
		}
		for(int i=1; i<=n; i++)
		{
			printf("%d ",b[i]);
		}
		puts("");
	}
}
