A. Unit Arraytime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputGiven an array a of length n, which elements are equal to -1 and 1. Let's call the array a good if the following conditions are held at the same time:  a_1 + a_2 + \ldots + a_n \ge 0;  a_1 \cdot a_2 \cdot \ldots \cdot a_n = 1. In one operation, you can select an arbitrary element of the array a_i and change its value to the opposite. In other words, if a_i = -1, you can assign the value to a_i := 1, and if a_i = 1, then assign the value to a_i := -1.Determine the minimum number of operations you need to perform to make the array a good. It can be shown that this is always possible.InputEach test consists of multiple test cases. The first line contains a single integer t (1 \le t \le 500) — the number of test cases. The description of the test cases follows.The first line of each test case contains a single integer n (1 \le n \le 100) — the length of the array a.The second line of each test case contains n integers a_1, a_2, \ldots, a_n (a_i = \pm 1) — the elements of the array a.OutputFor each test case, output a single integer — the minimum number of operations that need to be done to make the a array good.ExampleInput
74-1 -1 1 -15-1 -1 -1 1 14-1 1 -1 13-1 -1 -151 1 1 1 11-12-1 -1Output
1
1
0
3
0
1
2
NoteIn the first test case, we can assign the value a_1 := 1. Then a_1 + a_2 + a_3 + a_4 = 1 + (-1) + 1 + (-1) = 0 \ge 0 and a_1 \cdot a_2 \cdot a_3 \cdot a_4 = 1 \cdot (-1) \cdot 1 \cdot (-1) = 1. Thus, we performed 1 operation.In the second test case, we can assign a_1 := 1. Then a_1 + a_2 + a_3 + a_4 + a_5 = 1 + (-1) + (-1) + 1 + 1 = 1 \ge 0 and a_1 \cdot a_2 \cdot a_3 \cdot a_4 \cdot a_5 = 1 \cdot (-1) \cdot (-1) \cdot 1 \cdot 1 = 1. Thus, we performed 1 operation.In the third test case, a_1 + a_2 + a_3 + a_4 = (-1) + 1 + (-1) + 1 = 0 \ge 0 and a_1 \cdot a_2 \cdot a_3 \cdot a_4 = (-1) \cdot 1 \cdot (-1) \cdot 1 = 1. Thus, all conditions are already satisfied and no operations are needed.In the fourth test case, we can assign the values a_1 := 1, a_2 := 1, a_3 := 1. Then a_1 + a_2 + a_3 = 1 + 1 + 1 = 3 \ge 0 and a_1 \cdot a_2 \cdot a_3 = 1 \cdot 1 \cdot 1 = 1. Thus, we performed 3 operations.
74-1 -1 1 -15-1 -1 -1 1 14-1 1 -1 13-1 -1 -151 1 1 1 11-12-1 -1
1 1 0 3 0 1 2 
['greedy', 'math', '*800']