A. A.M. Deviationtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputA number a_2 is said to be the arithmetic mean of two numbers a_1 and a_3, if the following condition holds: a_1 + a_3 = 2\cdot a_2. We define an arithmetic mean deviation of three numbers a_1, a_2 and a_3 as follows: d(a_1, a_2, a_3) = |a_1 + a_3 - 2 \cdot a_2|.Arithmetic means a lot to Jeevan. He has three numbers a_1, a_2 and a_3 and he wants to minimize the arithmetic mean deviation d(a_1, a_2, a_3). To do so, he can perform the following operation any number of times (possibly zero):  Choose i, j from \{1, 2, 3\} such that i \ne j and increment a_i by 1 and decrement a_j by 1 Help Jeevan find out the minimum value of d(a_1, a_2, a_3) that can be obtained after applying the operation any number of times.InputThe first line contains a single integer t (1 \le t \le 5000)  — the number of test cases.The first and only line of each test case contains three integers a_1, a_2 and a_3 (1 \le a_1, a_2, a_3 \le 10^{8}).OutputFor each test case, output the minimum value of d(a_1, a_2, a_3) that can be obtained after applying the operation any number of times.ExampleInput
3
3 4 5
2 2 6
1 6 5
Output
0
1
0
NoteNote that after applying a few operations, the values of a_1, a_2 and a_3 may become negative.In the first test case, 4 is already the Arithmetic Mean of 3 and 5.d(3, 4, 5) = |3 + 5 - 2 \cdot 4| = 0In the second test case, we can apply the following operation:(2, 2, 6) \xrightarrow[\text{increment a_2}]{\text{decrement a_1}} (1, 3, 6)d(1, 3, 6) = |1 + 6 - 2 \cdot 3| = 1It can be proven that answer can not be improved any further.In the third test case, we can apply the following operations:(1, 6, 5) \xrightarrow[\text{increment a_3}]{\text{decrement a_2}} (1, 5, 6) \xrightarrow[\text{increment a_1}]{\text{decrement a_2}} (2, 4, 6)d(2, 4, 6) = |2 + 6 - 2 \cdot 4| = 0
3
3 4 5
2 2 6
1 6 5

0 1 0 
['math', 'number theory', '*800']