human
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.
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is Python code to find any such array b!
.

from sys import stdin
input=stdin.readline
rn=lambda:int(input())
rns=lambda:map(int,input().split())
rl=lambda:list(map(int,input().split()))
rs=lambda:input().strip()
YN=lambda x:print('YES') if x else print('NO')
mod=10**9+7

for _ in range(rn()):
    n=rn()
    a=rl()
    ans=[]
    def solve(b):
        ans=[]
        for i in range(0,len(b),2):
            ans.append(-b[i+1])
            ans.append(b[i])
        return ans
    if n%2==0:
        ans=solve(a)
    else:
        x,b,c=a[:3]
        l=len({x, b, c})
        if l==3:
            pre = [b-c,c-x,x-b]
        elif l==1:
            pre = [1,1,-2]
        else:
            if x==b:
                pre=[x,c-x,-x]
            elif x==c:
                pre=[x,-x,b-x]
            else:
                pre=[-b,b,x-b]
        ans=pre+solve(a[3:])
    print(*ans)