In a serious attempt to downsize (reduce) the dole queue, The New National Green Labour Rhinoceros Party has decided on the following strategy. Every day all dole applicants will be placed in a large circle, facing inwards. Someone is arbitrarily chosen as number 1, and the rest are numbered counter-clockwise up to N (who will be standing on 1's left). Starting from 1 and moving counter-clockwise, one labour official counts off k applicants, while another official starts from N and moves clockwise, counting m applicants. The two who are chosen are then sent off for retraining; if both officials pick the same person she (he) is sent off to become a politician. Each official then starts counting again at the next available person and the process continues until no-one is left. Note that the two victims (sorry, trainees) leave the ring simultaneously, so it is possible for one official to count a person already selected by the other official.

Input

Write a program that will successively read in (in that order) the three numbers (N, k and m; k, m > 0, 0 < N < 20) and determine the order in which the applicants are sent off for retraining. Each set of three numbers will be on a separate line and the end of data will be signalled by three zeroes (0 0 0).

For each triplet, output a single line of numbers specifying the order in which people are chosen. Each number should be in a field of 3 characters. For pairs of numbers list the person chosen by the counter-clockwise official first. Separate successive pairs (or singletons) by commas (but there should not be a trailing comma).

10 4 3
0 0 0

 4  8,  9  5,  3  1,  2  6,  10,  7

where  represents a space.

題意是n個人站成一圈，逆時針編號為1-n。有兩個官員，A從1開始逆時針數，B從n順時針數。在每一輪中，官員A數k個就停下來，官員B 數m個就停下來（注意有可能兩個官員停在同一個人上）。被官員選中的人離開隊伍

#include<stdio.h>
#define maxn 25
int n,k,m,a[maxn];
int go(int p,int d,int t)
{
	while(t--)
	{
		do{
			p=(p+d+n-1)%n+1;
		}while(a[p]==0);
	}
	return p;
}
int main()
{
	while(scanf("%d%d%d",&n,&k,&m)==3&&n)
	{
		for(int i=1;i<=n;i++)
		{
			a[i]=i;
		}
		int left=n;
		int p1=n,p2=1;
		while(left)
		{
			p1=go(p1,1,k);
			p2=go(p2,-1,m);
			printf("%3d",p1);
			left--;
			if(p2!=p1)
			{
				printf("%3d",p2);
				left--;
			}
			a[p1]=a[p2]=0;
			if(left)
			printf(",");
		}
		printf("\n");
	}
	return 0;
}