Competitive Programmer
Contest: Codeforces Global Round 6
Difficulty rating: 1000
Tags: chinese remainder theorem, math

Bob is a competitive programmer. He wants to become red, and for that he needs a strict training regime. He went to the annual meeting of grandmasters and asked $$$n$$$ of them how much effort they needed to reach red.

"Oh, I just spent $$$x_i$$$ hours solving problems", said the $$$i$$$-th of them.

Bob wants to train his math skills, so for each answer he wrote down the number of minutes ($$$60 \cdot x_i$$$), thanked the grandmasters and went home. Bob could write numbers with leading zeroes — for example, if some grandmaster answered that he had spent $$$2$$$ hours, Bob could write $$$000120$$$ instead of $$$120$$$.

Alice wanted to tease Bob and so she took the numbers Bob wrote down, and for each of them she did one of the following independently:

- rearranged its digits, or
- wrote a random number.

This way, Alice generated $$$n$$$ numbers, denoted $$$y_1$$$, ..., $$$y_n$$$.

For each of the numbers, help Bob determine whether $$$y_i$$$ can be a permutation of a number divisible by $$$60$$$ (possibly with leading zeroes).

Input format:
The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 418$$$) — the number of grandmasters Bob asked.

Then $$$n$$$ lines follow, the $$$i$$$-th of which contains a single integer $$$y_i$$$ — the number that Alice wrote down.

Each of these numbers has between $$$2$$$ and $$$100$$$ digits '0' through '9'. They can contain leading zeroes.

Output format:
Output $$$n$$$ lines.

For each $$$i$$$, output the following. If it is possible to rearrange the digits of $$$y_i$$$ such that the resulting number is divisible by $$$60$$$, output "red" (quotes for clarity). Otherwise, output "cyan".

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first example, there is one rearrangement that yields a number divisible by $$$60$$$, and that is $$$360$$$.

In the second example, there are two solutions. One is $$$060$$$ and the second is $$$600$$$.

In the third example, there are $$$6$$$ possible rearrangments: $$$025$$$, $$$052$$$, $$$205$$$, $$$250$$$, $$$502$$$, $$$520$$$. None of these numbers is divisible by $$$60$$$.

In the fourth example, there are $$$3$$$ rearrangements: $$$228$$$, $$$282$$$, $$$822$$$.

In the fifth example, none of the $$$24$$$ rearrangements result in a number divisible by $$$60$$$.

In the sixth example, note that $$$000\dots0$$$ is a valid solution.

Example input:
6
603
006
205
228
1053
0000000000000000000000000000000000000000000000

Example output:
red
red
cyan
cyan
cyan
red

Editorial:
Thanks to Chinese remainder theorem, a number is divisible by $$$60$$$ if and only if it is divisible by $$$3$$$ and $$$20$$$.
A number is divisible by $$$3$$$ if and only if the sum of its digits is divisible by $$$3$$$. Note that as the sum doesn't change if we reorder digits, it applies also to the sum of digits of $$$s$$$.
A number is divisible by $$$20$$$ if it ends in $$$20$$$, $$$40$$$, $$$60$$$, $$$80$$$ or $$$00$$$. Hence, it is necessary and sufficient if $$$s$$$ contains a $$$0$$$ and then at least one additional even digit.
Overall, there are three conditions to check:
• The digit sum is divisible by $$$3$$$.
• There is at least a single $$$0$$$.
• There are at least two even digits (including $$$0$$$s).