Distance and Axis
We have a point $$$A$$$ with coordinate $$$x = n$$$ on $$$OX$$$-axis. We'd like to find an integer point $$$B$$$ (also on $$$OX$$$-axis), such that the absolute difference between the distance from $$$O$$$ to $$$B$$$ and the distance from $$$A$$$ to $$$B$$$ is equal to $$$k$$$.  The description of the first test case. Since sometimes it's impossible to find such point $$$B$$$, we can, in one step, increase or decrease the coordinate of $$$A$$$ by $$$1$$$. What is the minimum number of steps we should do to make such point $$$B$$$ exist?
In the first test case (picture above), if we set the coordinate of $$$B$$$ as $$$2$$$ then the absolute difference will be equal to $$$|(2 - 0) - (4 - 2)| = 0$$$ and we don't have to move $$$A$$$. So the answer is $$$0$$$.In the second test case, we can increase the coordinate of $$$A$$$ by $$$3$$$ and set the coordinate of $$$B$$$ as $$$0$$$ or $$$8$$$. The absolute difference will be equal to $$$|8 - 0| = 8$$$, so the answer is $$$3$$$.  
The first line contains one integer $$$t$$$ ($$$1 \le t \le 6000$$$) — the number of test cases.The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$0 \le n, k \le 10^6$$$) — the initial position of point $$$A$$$ and desirable absolute difference.
For each test case, print the minimum number of steps to make point $$$B$$$ exist.
for l in [*open(0)][1:]:
    (N, K) = map(int, l.split())
    print(max(K - N, K + N & 1))
import math

def iamdumb(a, k):
    if a == k:
        return 0
    elif a > k:
        if a % 2 == 0 and k % 2 == 0 or (a % 2 != 0 and k % 2 != 0):
            return 0
        return 1
    else:
        return k - a
t = int(input())
result = []
for j in range(t):
    m = input()
    alist = [int(f) for f in m.split()]
    result.append(iamdumb(alist[0], alist[1]))
for res in result:
    print(res)
for l in [*open(0)][1:]:
    (N, K) = map(int, l.split())
    print(max(K - N, K + N & 1))