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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 gcd(a, b):
    r = a % b
    while r:
        a = b
        b = r
        r = a % b
    return b

def lcm(a, b):
    return abs(a*b) // gcd(a, b)

t = int(input())
for test_case in range(t):
    n = int(input())
    a = [int(i) for i in input().split()]
    b = []
    if n % 2:
        sum_of_first_two = a[0] + a[1]
        num_multiplier = 1
        if not sum_of_first_two:
            num_multiplier = 2
            sum_of_first_two = a[0] + a[1]*num_multiplier
        lowest_common_multiple = lcm(sum_of_first_two, a[2])
        b.append(f"{lowest_common_multiple / sum_of_first_two:.0f}")
        b.append(f"{lowest_common_multiple / sum_of_first_two * num_multiplier:.0f}")
        b.append(f"{lowest_common_multiple / a[2] * -1:.0f}")

        for i in range(3, n, 2):
            lowest_common_multiple = lcm(a[i], a[i+1])
            b.append(f"{lowest_common_multiple / a[i]:.0f}")
            b.append(f"{lowest_common_multiple / a[i+1] * -1:.0f}")
    else:
        for i in range(0, n, 2):
            lowest_common_multiple = lcm(a[i], a[i+1])
            b.append(f"{lowest_common_multiple / a[i]:.0f}")
            b.append(f"{lowest_common_multiple / a[i+1] * -1:.0f}")

    print(" ".join(b))