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.

Input

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 $$$a_3$$$ $$$(1 \le a_1, a_2, a_3 \le 10^{8})$$$.

Output

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

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$$$}}$$$ $$$(2, 4, 6)$$$

$$$d(2, 4, 6) = |2 + 6 - 2 \cdot 4| = 0$$$

A.M. Deviation

Examples