Solve the programming task below in a Python markdown code block.
Read problem statements in [Russian] and [Mandarin Chinese].

You are given two integers A, B. You have to choose an integer X in the range [minimum(A, B), maximum(A, B)] such that: 

\Big \lceil \frac{B - X}{2} \Big \rceil + \Big \lceil \frac{X - A}{2} \Big \rceil is maximum. 

What is the maximum sum you can obtain if you choose X optimally?

Note: \lceil x \rceil : Returns the smallest integer that is greater than or equal to x (i.e  rounds up to the nearest integer). For example, \lceil 1.4 \rceil = 2, \lceil 5 \rceil = 5, \lceil -1.5 \rceil = -1, \lceil -3 \rceil = -3 , \lceil 0 \rceil = 0.

------ Input Format ------ 

- First line will contain T, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, two integers A, B.

------ Output Format ------ 

For each testcase, output the maximum sum of \lceil \frac{B - X}{2} \rceil + \lceil \frac{X - A}{2} \rceil.

------ Constraints ------ 

$1 ≤ T ≤ 10^{5}$
$1 ≤ A, B ≤ 10^{9}$

----- Sample Input 1 ------ 
3
2 2
1 11
13 6

----- Sample Output 1 ------ 
0
6
-3
----- explanation 1 ------ 
- In the first test case, there is only one possible value of $X$ which is equal to $2$. So the sum is equal to $\lceil \frac{2 - 2}{2} \rceil + \lceil \frac{2 - 2}{2} \rceil$ = $\lceil \ 0 \rceil + \lceil \ 0 \rceil$  = $0 + 0 = 0$.
   
- In the second test case, we can choose $X$ to be $2$. So sum is equal to $\lceil \frac{11 - 2}{2} \rceil + \lceil \frac{2 - 1}{2} \rceil$ = $\lceil \ 4.5 \rceil + \lceil \ 0.5 \rceil$ = $5 + 1 = 6$. There is no possible way to choose an integer $X$ in the range $[1, 11]$ such that sum is greater than $6$.