Divide and Conquer
An array $$$b$$$ is good if the sum of elements of $$$b$$$ is even. You are given an array $$$a$$$ consisting of $$$n$$$ positive integers. In one operation, you can select an index $$$i$$$ and change $$$a_i := \lfloor \frac{a_i}{2} \rfloor$$$. $$$^\dagger$$$Find the minimum number of operations (possibly $$$0$$$) needed to make $$$a$$$ good. It can be proven that it is always possible to make $$$a$$$ good.$$$^\dagger$$$ $$$\lfloor x \rfloor$$$ denotes the floor function — the largest integer less than or equal to $$$x$$$. For example, $$$\lfloor 2.7 \rfloor = 2$$$, $$$\lfloor \pi \rfloor = 3$$$ and $$$\lfloor 5 \rfloor =5$$$.
In the first test case, array $$$a$$$ is already good.In the second test case, we can perform on index $$$2$$$ twice. After the first operation, array $$$a$$$ becomes $$$[7,2]$$$. After performing on index $$$2$$$ again, $$$a$$$ becomes $$$[7,1]$$$, which is good. It can be proved that it is not possible to make $$$a$$$ good in less number of operations.In the third test case, $$$a$$$ becomes $$$[0,2,4]$$$ if we perform the operation on index $$$1$$$ once. As $$$[0,2,4]$$$ is good, answer is $$$1$$$.In the fourth test case, we need to perform the operation on index $$$1$$$ four times. After all operations, $$$a$$$ becomes $$$[0]$$$. It can be proved that it is not possible to make $$$a$$$ good in less number of operations.
Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. The description of the test cases follows.The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 50$$$) — the length of the array $$$a$$$.The second line of each test case contains $$$n$$$ space-separated integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \leq a_i \leq 10^6$$$) — representing the array $$$a$$$.Do note that the sum of $$$n$$$ over all test cases is not bounded.
For each test case, output the minimum number of operations needed to make $$$a$$$ good.
for s in [*open(0)][2::2]:
    a = (*map(int, s.split()),)
    print(sum(a) % 2 * min((len(x) - len(x.rstrip(x[-1])) for x in map(bin, a))))
for s in [*open(0)][2::2]:
    a = (*map(int, s.split()),)
    print(sum(a) % 2 * min((len(x) - len(x.rstrip(x[-1])) for x in map(bin, a))))
for sdr in [*open(0)][2::2]:
    p = (*map(int, sdr.split()),)
    print(sum(p) % 2 * min((len(x) - len(x.rstrip(x[-1])) for x in map(bin, p))))