B'A 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): 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. The first line contains a single integer t (1 <= t <= 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 <= a_1, a_2, a_3 <= 10^{8}) . For 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. Note 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| = 0 In 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| = 1 It 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 '...