F.A.Q
Hand In Hand
Online Acmers
Problem Archive
Realtime Judge Status
Authors Ranklist
 
     C/C++/Java Exams     
ACM Steps
Go to Job
Contest LiveCast
ICPC@China
Best Coder beta
VIP | STD Contests
    DIY | Web-DIY beta
Author ID 
Password 
 Register new ID

Have meal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1807    Accepted Submission(s): 1110


Problem Description
I have been in school for several years, so I have visited all messes here. Now I have lost intersts in all of the foods. So when during the meal time, I don¡¯t know which mess I should go to. So I came up with a solution.
There are 4 messes in our school, I number them from 0 to 3. Then I says ¡°Big Bing Small Jiang, Point Who Is Who!¡±, when I say the first word I point to the mess which is numbered 0, when I say the i-th (i>1) word I point to the mess whose number is one larger than the previous one. In case of the number of previous mess is 3, I will point to 0 again. I will go to the mess which I point to last time. Thus in this case I will go to the mess which is numbered 3. The following table explains the course of my solution to this case.
$\begin{array} {|c|c|c|c|}
\hline
\quad \text{ Word I say} \quad &\quad \text{ Big} \quad &\quad \text{ Bing} \quad &\quad \text{ Small} \quad &\quad \text{ Jiang} \quad &\quad \text{ Point} \quad &\quad \text{ Who} \quad &\quad \text{ Is} \quad &\quad \text{ Who} \quad\\
\hline
\hline
\quad \text{ Mess id I point to} \quad &\quad \text{ 0} \quad &\quad \text{ 1} \quad &\quad \text{ 2} \quad &\quad \text{ 3} \quad &\quad \text{ 0} \quad &\quad \text{ 1} \quad &\quad \text{ 2} \quad &\quad \text{ 3} \quad\\
\hline

\end{array}\\
$
I will go to university after several days, I have heard that there are so many messes in it. So I will apply my solution again. Surpose there are n messes which are numberd through 0 to n-1, and I will say m words. When I say the first word I point to the mess which is numbered 0, when I say the i-th (i>1) word I point to the mess whose number is one larger than the previous one. In case of the number of previous mess is n-1, I will point to 0 again. I will go to the mess which I point to last time. So which mess will I point to?.
It is so time-consuming to count it through manual work. So I want you to write a program to help me. Would you help me?
 

Input
Multi test cases (about 10000), every case contain two integers n and m in a single line.

[Technical Specification]
1<=n, m<=100
 

Output
For each case, output the number of the mess which I should go to.
 

Sample Input
4 3 1 100
 

Sample Output
2 0
 

Source
 

Statistic | Submit | Discuss | Note
Hangzhou Dianzi University Online Judge 3.0
Copyright © 2005-2024 HDU ACM Team. All Rights Reserved.
Designer & Developer : Wang Rongtao LinLe GaoJie GanLu
Total 0.000000(s) query 1, Server time : 2024-11-26 00:15:53, Gzip enabled