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.

#include<stdio.h>
#include<string.h>
int main()
{
	int *p,*q,i,j,n,k,m,a[25]={},tot=0,t1,t2,k1,k2,case1,case2,c1=0,c2=0;
	while(scanf("%d%d%d",&n,&k,&m)!=EOF&&n&&k&&m)
	{
	tot=0;
	memset(a,0,sizeof(a));
	for(i=n,j=1;i>=1;i--,j++)
	{
		a[i]=j;
	}
	
	q=a+1;p=a+n;
	t1=*p;t2=*q;
	while(tot!=n)
	{		k1=k;
			k2=m;
			c1=c2=0;
		if(*p==0)
		c1--;
		while(c1!=k-1)
		{	
			p--;
			if(p<=a)
			{
				p=a+n;
			}
			if(*p!=0)
			{
				c1++;
			}
			t1=*p;
		}
		if(*q==0)
		c2--;
		while(c2!=m-1)
		{
			q++;
			if(q>=a+n+1)
			{
				q=a+1;
			}
			if(*q!=0)
			{
				c2++;
			}
			t2=*q;
		}
		
		if(*p==*q)
		{
			tot++;
			if(tot==n-1&&tot==n)
			printf("%3d",*p);
			else
			printf("%3d",*p);
			*p=*q=0;
			p--;
			if(p<=a)
			p=a+n;
			q++;
			if(q>=a+n)
			q=a+1;
			
		}
		else
		{
			tot+=2;
			if(tot==n-1||tot==n)
			printf("%3d%3d",*p,*q);
			else
			printf("%3d%3d",*p,*q);
				*p=*q=0;
				p--;
				if(p<=a)
				p=a+n;
				q++;
				if(q>=a+n)
				q=a+1;
				
		}
			if(tot!=n)
			printf(",");
	}
	if(tot!=n)
	printf(",");
	printf("\n");
	}
}