Solve the programming task below in a Python markdown code block.
For a positive integer M, MoEngage defines \texttt{digitSum(M)} as the sum of digits of the number M (when written in decimal). 

For example, \texttt{digitSum(1023)} = 1 + 0 + 2 + 3 = 6.

Given a positive integer N, find the smallest integer X strictly greater than N such that:
\texttt{digitSum(N)} and \texttt{digitSum(X)} have different [parity], i.e. one of them is odd and the other is even.

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

- The first line contains an integer T, the number of test cases. The description of the T test cases follow.
- Each test case consists of a single line of input with a single integer, the number N.

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

- For each test case, print in a single line, an integer, the answer to the problem.

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

$1 ≤ T ≤ 1000$
$1 ≤ N < 10^{9}$

----- Sample Input 1 ------ 
3
123
19
509
----- Sample Output 1 ------ 
124
21
511
----- explanation 1 ------ 
Test Case $1$: $\texttt{digitSum}(123) = 1 + 2 + 3 = 6$ is even and $\texttt{digitSum}(124) = 1 + 2 + 4 = 7$ is odd, so the answer is $124$.

Test Case $2$: $\texttt{digitSum}(19) = 1 + 9 = 10$ is even, $\texttt{digitSum}(20) = 2 + 0 = 2$ is also even, whereas $\texttt{digitSum}(21) = 2 + 1 = 3$ is odd. Hence, the answer is $21$.

Test Case $3$: $\texttt{digitSum}(509) = 5 + 0 + 9 = 14$ is even, $\texttt{digitSum}(510) = 5 + 1 + 0 = 6$ is also even, whereas $\texttt{digitSum}(511) = 5 + 1 + 1 = 7$ is odd. Hence, the answer is $511$.