A. Unit Array
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Given 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.

Input

Each 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 $$$a$$$.

Output

For each test case, output a single integer — the minimum number of operations that need to be done to make the $$$a$$$ array good.

Example
Input
7
4
-1 -1 1 -1
5
-1 -1 -1 1 1
4
-1 1 -1 1
3
-1 -1 -1
5
1 1 1 1 1
1
-1
2
-1 -1
Output
1
1
0
3
0
1
2
Note

In 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.