Intruder Outsmarting

The security key has $W$ wheels in a row. Each wheel has the numbers $1$ through $N$ printed in order. One rotation moves the number shown on a wheel to the next or the previous number. Numbers wrap around, so the number after $N$ is $1$, and the number before $1$ is $N$.

There is no hidden password. To activate the key, the numbers shown must form a palindrome, meaning the sequence reads the same from left to right and from right to left. To slow down intruders, the wheels only turn in steps of $D$. In one operation, a wheel showing $x$ can be changed to show $x - D$ or $x + D$. The wraparound works like this: if $x - D < 1$, the number shown becomes $x - D + N$, and if $x + D > N$, the number shown becomes $x + D - N$.

Given the current number on each wheel, find the minimum number of operations needed to make the sequence of shown numbers a palindrome, or report that it is impossible.

The first line contains the number of test cases $T$. Each test case consists of two lines. The first line has three integers $W$, $N$, and $D$. The second line has $W$ integers $X_1, X_2, \dots, X_W$, where $X_i$ is the number currently shown on the $i$-th wheel from the left.

For each test case, output one line containing Case #x: y, where $x$ is the test case number starting from 1, and $y$ is the minimum number of operations needed to make the shown sequence a palindrome. If it is impossible, output IMPOSSIBLE instead of $y$.

• $1 \le T \le 100$.
• $1 \le D \le N - 1$.
• $1 \le X_i \le N$, for all $i$.

In Sample Case #1, the sequence can be made $5, 4, 5, 4, 5$, which is a palindrome, with $3$ operations: one addition on the first and fourth wheels, and one subtraction on the fifth wheel. No sequence of fewer moves works.

In Sample Case #2, the sequence is already a palindrome, so no operations are needed.

In Sample Case #3, both numbers must be equal for the sequence to be a palindrome. The wheel values only move by $2$, and the two current numbers have different parity, so this cannot be done.