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
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
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”.Exampleinput
6 603 006 205 228 1053 0000000000000000000000000000000000000000000000
output
red red cyan cyan cyan red
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.
Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
while (n--) {
string s;
cin >> s;
int sum = 0;
bool any_even = false;
bool any_zero = false;
for (char c : s) {
sum += (int) (c - '0');
if (c == '0') {
if (any_zero) {
any_even = true;
} else {
any_zero = true;
}
} else {
if (c % 2 == 0) {
any_even = true;
}
}
}
cout << (any_even && any_zero && sum % 3 == 0 ? "red" : "cyan") << '\n';
}
return 0;
}
