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$ curl repovive.com/contests/8/problems/a

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$ curl repovive.com/contests/8/problems/a

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8A. Make it Average

500 ptstype :zen to enter zen mode

by AmirMohammad Shahrezaei (AmShZ)

Time: 2sMemory: 256 MB

You are given three integers $a$ , $b$ , and $c$ .

In one operation, you may choose one of the three numbers and either increase it by $1$ or decrease it by $1$ .

Find the minimum number of operations needed to make $b$ equal to the arithmetic mean of $a$ and $c$ .

In other words, after the operations, the condition $\frac{a+c}{2}=b$ must hold.

Constraints

$1 \le t \le 10^4$
$1 \le a,b,c \le 10^6$

Input

t

\left. \begin{array}{l} a \quad b \quad c \end{array} \right\} \times t

Output

\left. \begin{array}{l} ans \end{array} \right\} \times t

Samples

Input

Case 1

1 2 3

Case 2

1 2 4

Case 3

10 1 10

Case 4

5 10 5

Case 5

7 4 10

Explanation

Case 1

The number $2$ is already the arithmetic mean of $1$ and $3$ , so no operation is needed.

Case 2

If we decrease $c$ by $1$ , the numbers become $1,2,3$ , and $2$ is the arithmetic mean of $1$ and $3$ .

Case 3

We can increase $b$ from $1$ to $10$ using $9$ operations. Then $b$ becomes the arithmetic mean of $10$ and $10$ .

Case 4

We can decrease $b$ from $10$ to $5$ using $5$ operations. Then $b$ becomes the arithmetic mean of $5$ and $5$ .

Case 5

For example, we can increase $b$ from $4$ to $8$ using $4$ operations. Then the numbers are $7,8,10$ . After decreasing $c$ by $1$ , the numbers become $7,8,9$ , and $8$ is the arithmetic mean of $7$ and $9$ .

Output

Case 1

Case 2

Case 3

Case 4

Case 5

You are given three integers $a$ , $b$ , and $c$ .

In one operation, you may choose one of the three numbers and either increase it by $1$ or decrease it by $1$ .

Find the minimum number of operations needed to make $b$ equal to the arithmetic mean of $a$ and $c$ .

In other words, after the operations, the condition $\frac{a+c}{2}=b$ must hold.

$1 \le t \le 10^4$
$1 \le a,b,c \le 10^6$

t

\left. \begin{array}{l} a \quad b \quad c \end{array} \right\} \times t

\left. \begin{array}{l} ans \end{array} \right\} \times t

The number $2$ is already the arithmetic mean of $1$ and $3$ , so no operation is needed.

If we decrease $c$ by $1$ , the numbers become $1,2,3$ , and $2$ is the arithmetic mean of $1$ and $3$ .

We can increase $b$ from $1$ to $10$ using $9$ operations. Then $b$ becomes the arithmetic mean of $10$ and $10$ .

We can decrease $b$ from $10$ to $5$ using $5$ operations. Then $b$ becomes the arithmetic mean of $5$ and $5$ .