Kattis

# Invoker (Hard)

You are an Arch-Magus Invoker, one of the strongest sorcerers that is able to command sources of magic.

The difference between an Arch-Magus Invoker and a regular Invoker is that an Arch-Magus holds an advanced degree or is pursuing a post-doc in mathemagical studies at the Institute of Arcane Studies.

## Magical Training

There are two sources of magic available to invokers to control, $\mathbb {R}$eagus and $\mathbb {I}$magus. Each source has a potency coefficient in front of it. This potency is a real number at most $10^6$ in absolute value.

As an invoker you have the choice of combining two sources of magic (denoted $\oplus$), or casting one on to another (denoted $\otimes$).

The rules for combining sources is as follows:

• Combining a source of Reagus or Imagus with another source of equal type creates the same type but you add the potencies.

• $x\mathbb {R}\oplus y\mathbb {R}=(x+y)\mathbb {R}$

• $x\mathbb {I}\oplus y\mathbb {I}=(x+y)\mathbb {I}$

• Combining a source of Reagus and Imagus together creates a new complex form of magic.

• $x\mathbb {R}\oplus y\mathbb {I} \in \mathbb {C}_{\mathbb {M}}$

Casting sources of magic has some different laws:

• Casting Reagus on to Reagus creates a source of Reagus whose potency is the product of the previous sources.

• $x\mathbb {R}\otimes y\mathbb {R} = (x\cdot y)\mathbb {R}$

• Casting Imagus on to another source of Imagus creates anti-Reagus (negative potency) with a potency of the product of the two sources.

• $x\mathbb {I}\otimes y\mathbb {I} = -(x\cdot y)\mathbb {R}$

• Casting Reagus on Imagus or Imagus on Reagus creates Imagus with a potency of the product of the two sources.

• $x\mathbb {R}\otimes y\mathbb {I} = (x\cdot y)\mathbb {I}$

Additionally, every magic source $s$ has a magic complement $s^\dagger$ such that:

\begin{aligned} s=x\mathbb {R}& \rightarrow s^\dagger =s\\ s=y\mathbb {I}& \rightarrow s^\dagger =-s\\ s=x\mathbb {R}\oplus y\mathbb {I}& \rightarrow s+s^\dagger =2x\mathbb {R} \end{aligned}

Casting a complex magic source on its complement (or visa versa), generates a non-negative number known as the potency norm of the source.

$s\otimes s^\dagger = |s|^2$

As an Arch-Magus you are well studied in an additional source of magic: $\mathbb {J}$ae.

Jae acts like Imagus when near the presence of Reagus. However, when Jae is cast on Imagus it can form a $4$th form of magic: $\mathbb {K}$aos!!

Casting either form of Imagus, Jae, or Kaos on itself creates anti-Reagus.

$\mathbb {I}\otimes \mathbb {I} = \mathbb {J}\otimes \mathbb {J} = \mathbb {K}\otimes \mathbb {K} = -\mathbb {R}$

Casting Imagus on to Jae actually creates anti-Kaos however $-\mathbb {K}$.

The complete casting table between the four elements is as follows:

$\begin{array}{l|rrrr} \bigotimes & \mathbb {R} & \mathbb {I} & \mathbb {J} & \mathbb {K} \\ \hline \mathbb {R} & \mathbb {R} & \mathbb {I} & \mathbb {J} & \mathbb {K} \\ \mathbb {I} & \mathbb {I} & -\mathbb {R} & \mathbb {K} & -\mathbb {J} \\ \mathbb {J} & \mathbb {J} & -\mathbb {K} & -\mathbb {R} & \mathbb {I} \\ \mathbb {K} & \mathbb {K} & \mathbb {J} & -\mathbb {I} & -\mathbb {R} \end{array}$

The same rules for combination exist for Jae and Kaos as they do for Imagus.

If you haven’t noticed, this means that casting is non-commutative. That is:

$a\otimes b \neq b\otimes a$

where $a\otimes b$ is called a right cast from $b$ to $a$, and $b\otimes a$ is called a left cast from $b$ to $a$.

Complex combinations and castings are allowed such as:

$\left(a\mathbb {R}\oplus b\mathbb {K}\right)\otimes \left(c\mathbb {I}\oplus d\mathbb {J}\oplus e\mathbb {K}\right)$

Casting is distributive, which means $s\otimes (a\oplus b) = (s\otimes a) \oplus (s\otimes b)$.

For every source there is also guaranteed to exist an inverse source that when cast will neutralize the magical energies.

$\forall s\in \mathbb {M}~ \exists s^{-1}\in \mathbb {M} : s\otimes s^{-1} = \epsilon$

where $\mathbb {M}$ is the set of non-zero magical sources and $\epsilon$ is the neutral source, or $1\mathbb {R}$.

Complements for Jae and Kaos work the same way they do for Imagus.

### Problem

You are currently locked in battle against an equally powerful foe. You recognize that you can defeat them by neutralizing their source of magic. That is, you cast a spell on them that will turn their source into the neutral source $\epsilon = 1\mathbb {R}$.

However, magic is not without its consequences. Any source you cast on your opponent will also be cast on you as well, changing your current magical source.

When you cast a spell on your opponent (and simultaneously on yourself) the magical cast is a right cast (described above), that is:

$x\otimes y$

where $x$ is the initial source and $y$ is the source you have cast.

When the dust settles and you are declared victorious, what is your new magical source?

## Input

The inputs are two magical sources separated by new lines.

The first line contains your source: $a\mathbb {R}\oplus b\mathbb {I}\oplus c\mathbb {J}\oplus d\mathbb {K}$. The second line contains your opponent’s source: $e\mathbb {R}\oplus f\mathbb {I}\oplus g\mathbb {J}\oplus h\mathbb {K}$.

All sources are integers in the range $-10^6 \le a,b,c,d,e,f,g,h \le 10^6$, and neither source has all four potencies equal to zero.

## Output

Output your resulting source after you have annihilated your opponent. The output should be four space separated values $x,y,z,w$. $x$ is the potency of Reagus, $y$ the potency of Imagus, $z$ the potency of Jae, and $w$ the potency of Kaos present in your final source.

Your answers will be accepted if they have an absolute or relative error of at most $10^{-5}$.

Sample Input 1 Sample Output 1
1 2 3 4
1 2 3 4

1.0 0.0 0.0 0.0

Sample Input 2 Sample Output 2
0 1 0 0
0 0 1 0

0.0 0.0 0.0 -1.0