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What is a Python program to find any such array b! given the following:
PYTHON3 solution for "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.":
assistant
#Here is Python program to find any such array b!
.
 def solve_case():
    n = int(input())
    a = list(map(int, input().split()))
    b = [0] * n
    i = -2
    j = -1
    if n % 2 == 0:
        for p in range(n // 2):
            i = 2 * p
            j = 2 * p + 1
            b[i] = a[j]
            b[j] = -a[i]
    else:
        for p in range(n // 2 - 1):
            i = 2 * p
            j = 2 * p + 1
            b[i] = a[j]
            b[j] = -a[i]
        i += 2
        p = -1
        q = -1
        # i, i+1, i+2
        if a[i] + a[i + 1] != 0:
            p = i
            q = i + 1
            r = i + 2
        elif a[i] + a[i + 2] != 0:
            p = i
            q = i + 2
            r = i + 1
        else:  # a[i+1] + a[i+2] != 0 (by def)
            p = i + 1
            q = i + 2
            r = i
        b[r] = (a[p] + a[q])
        b[p] = b[q] = -a[r]
    # s = sum(b[i] * a[i] for i in range(n))
    # print(f'sum = {s}')
    print(' '.join(map(str, b)))

T = int(input())
for t in range(T):
    solve_case()