CF1605A A.M. Deviation

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): - 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.

The 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}) $ .

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 $