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.
import sys
import io, os
input = sys.stdin.buffer.readline
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

from collections import defaultdict

t = int(input())
for _ in range(t):
    n = int(input())
    A = list(map(int, input().split()))
    C = []
    for i, a in enumerate(A):
        C.append((abs(a), i))
    C.sort(key=lambda x: x[0])
    B = [0]*n
    k = C[-1][1]
    c = A[k]
    D = []
    for i, a in enumerate(A):
        if i != k:
            D.append((a, i))
    D.sort(key=lambda x: x[0])
    x = 0
    for i in range(n-1):
        _, j = D[i]
        a = A[j]
        if i%2 == 0:
            B[j] = c
        else:
            B[j] = -c
        x += a*B[j]
    B[k] = -x//c
    print(*B)
    #ans = 0
    #for a,b in zip(A, B):
        #ans += a*b
    #print(ans)
