A. Competitive Programmer
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).
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 \(n\) 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"".
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.