Esspe-Peasee

Esspe-Peasee is an ancient game played by children throughout the land of Acmania. The rules are simple.

A player quibs the yorba at the kwonk. If the yorba hurms the kwonk, the player earns a foom; if the yorba hurfs the kwonk, the player earns a foob. The objective is to assemble a twob using as few quibs as possible.

Every group of children has its own opinion about the value of a foom, the value of a foob, and the value of a twob. Everyone agrees, however, that a foob is worth more than a foom, and that a twob is worth more than a foob. A foom and a foob each fit in a 32-bit integer, and a twob fits in a 64-bit integer.

Each quib yields exactly one foom or one foob, so if you collect $A$ fooms and $B$ foobs you have spent $A + B$ quibs and accumulated a value of $A \cdot (\text{foom}) + B \cdot (\text{foob})$. You want this total to equal the value of a twob while using the fewest quibs, that is, while minimising $A + B$.

You will be given several game instances. Each instance is a single line with three non-negative integers: the value of a foom, the value of a foob, and the value of a twob, in that order. Input ends with a line containing three zeros (0 0 0), which must not be processed.

For each instance, print a single line A fooms and B foobs for a twob!, where $A$ fooms plus $B$ foobs add up to exactly one twob and $A + B$ is as small as possible. Pluralise the words correctly: write foom / foob when the count is exactly $1$, and fooms / foobs otherwise (including a count of $0$).

If no such pair of non-negative counts exists, print the age-old chant instead: Unquibable!