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.
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$$$}}$$$ $$$(2, 4, 6)$$$$$$d(2, 4, 6) = |2 + 6 - 2 \cdot 4| = 0$$$
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.
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def i():
    s = list(map(int, input().split()))
    c = abs(s[0] + s[2] - 2 * s[1]) % 3
    return ['1', '0'][c == 0]
t = int(input())
a = []
for _ in range(t):
    a.append(i())
print('\n'.join(a))
for k in [*open(0)][1:]:
    print('011'[sum(map(int, k.split())) % 3])
class Solution:

    def __init__(self):
        pass

    def solve(self):
        t = int(input())
        result = []
        for i in range(t):
            (a, b, c) = map(int, input().split())
            if abs(a + c - 2 * b) % 3 == 0:
                result.append(0)
            else:
                result.append(1)
        for x in result:
            print(x)
if __name__ == '__main__':
    sol = Solution()
    sol.solve()