Codeforces Round 125 (Div. 2)
A
Hexadecimal's theorem
Codeforces
900
原题
←B →
翻译中…

Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.

翻译中…

Let's remember how Fibonacci numbers can be calculated. $*F*_{0} = 0$, $*F*_{1} = 1$, and all the next numbers are $*F*_{*i*} = *F*_{*i* - 2} + *F*_{*i* - 1}$.

翻译中…

So, Fibonacci numbers make a sequence of numbers: $0$, $1$, $1$, $2$, $3$, $5$, $8$, $13$, ...

翻译中…

If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number $*n*$ by three not necessary different Fibonacci numbers or say that it is impossible.

翻译中…

Input

翻译中…

The input contains of a single integer $*n*$ $(0 ≤ *n* < 10^{9})$ — the number that should be represented by the rules described above. It is guaranteed that $*n*$ is a Fibonacci number.

翻译中…

Output

翻译中…

Output three required numbers: $*a*$, $*b*$ and $*c*$. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.

翻译中…

If there are multiple answers, print any of them.

翻译中…

样例

Input #1
3
Output #1
1 1 1
InputCopy OutputCopy
Input #2
13
Output #2
2 3 8
InputCopy OutputCopy
题目信息
题号
199A
时间限制
2 s
内存限制
256 MB
标签
暴力
构造
模拟
数论

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。