Venn Diagram

Alice teaches privately, and preparing the material is part of her job. Right now she is drawing a Venn diagram of two sets $A$ and $B$.

A Venn diagram shows how sets relate to one another. In a Venn diagram of $A$ and $B$, the outer rectangle is the universal set $U$, and the two circles inside it are $A$ and $B$. The region where the circles overlap corresponds to the intersection $A \cap B$.

Alice wants the picture to carry the numbers too, so she imposed a condition: the area of every region must equal the number of elements of the set that region stands for. One circle has area $|A|$, the other has area $|B|$, and the overlap of the two circles has area $|A \cap B|$. Here $|X|$ is the number of elements of a set $X$.

The rectangle is already drawn. Its lower left corner is $(0, 0)$ and its upper right corner is $(U_W, U_H)$. Both circles have to lie inside that rectangle. Write a program that computes the centre and the radius of each circle.

The input holds several datasets. There are at most $300$ of them. Each dataset is one line with five integers.

UW UH |A| |B| |A∩B|

$U_W$ and $U_H$ ($1 \le U_W, U_H \le 100$) are the width and the height of the rectangle that stands for the universal set $U$. $|A|$, $|B|$ and $|A \cap B|$ ($1 \le |A|, |B| \le 10000$, $0 \le |A \cap B| \le \min(|A|, |B|)$) are the numbers of elements of $A$, $B$ and $A \cap B$. The last line holds five zeroes and is not a dataset.

Whether the two circles can be drawn does not change when $U_W$ and $U_H$ move by up to $0.01$ in either direction.

Print one line for each dataset. Many placements satisfy Alice's condition, so print the single placement that the rule below picks out.

The radii follow from the areas: $R_A = \sqrt{|A| / \pi}$ and $R_B = \sqrt{|B| / \pi}$.

Let $d$ be the smallest distance between the two centres at which the overlap area equals $|A \cap B|$. If $|A \cap B| = 0$, then $d = R_A + R_B$. If $|A \cap B| = \min(|A|, |B|)$, then $d = 0$. Otherwise exactly one $d$ with $|R_A - R_B| < d < R_A + R_B$ gives that overlap area.

The two circles can be drawn exactly when both of the following hold.

• $2\max(R_A, R_B) \le \min(U_W, U_H)$
• $d \le D$, where $D = \sqrt{(U_W - R_A - R_B)^2 + (U_H - R_A - R_B)^2}$

If either one fails, print impossible.

Otherwise let $s = d / D$, and let $s = 0$ when $D = 0$. Print the six values

• $X_A = U_W/2 - s (U_W/2 - R_A)$
• $Y_A = U_H/2 - s (U_H/2 - R_A)$
• $R_A$
• $X_B = U_W/2 + s (U_W/2 - R_B)$
• $Y_B = U_H/2 + s (U_H/2 - R_B)$
• $R_B$

in that order, on one line, separated by single spaces, each with exactly nine digits after the decimal point. In words: the centre of circle $A$ lies on the segment from the centre of the rectangle to the point $(R_A, R_A)$, the centre of circle $B$ lies on the segment from the centre of the rectangle to the point $(U_W - R_B, U_H - R_B)$, and both centres sit the same fraction $s$ of the way along their own segment.