#include <cstdio>

#define NMAX 1000005
struct interval {int left, right, color;};

int N,A,B,C;
interval I[NMAX];
int color[NMAX],next[NMAX],father[NMAX];

inline int min(int x, int y)
{
	return (x<y)?x:y;
}

inline int max(int x, int y)
{
	return (x>y)?x:y;
}

void init()
{
	I[1].left = min(A,B);
	I[1].right = max(A,B);
	I[1].color = C;
	for(int i=2;i<N;i++)
	{
		A = (long long) A * i % (long long) N;
		B = (long long) B * i % (long long) N;
		C = (long long) C * i % (long long) N;
		I[i].left = min(A,B);
		I[i].right = max(A,B);
		I[i].color = C;
	}
	for(int i=1;i<=N;i++)
	{
		color[i] = 0;
		next[i] = i;
		father[i] = -1;
	}
}

inline int root(int u)
{
	if(father[u]<0)
		return u;
	else
	{
//		father[u]=root(father[u]);
//		return father[u];
		return root(father[u]);
	}
}

inline void join(int u, int v) // u si v sunt 2 radacini de arbori
{
	if(father[u]<father[v]) // u mai adanc decat v
		father[v]=u;
	else
		if(father[u]>father[v]) // v mai adanc decat u
			father[u]=v;
		else // aceeasi adancime
		{
			father[u]=v;
			father[v]--;
		}
}

void solve()
{
	int u,v;
	for(int k=N-1; k>=1; --k)
	{
		for(int i=I[k].left; i<=I[k].right;)
		{
			u=root(i);
			v=root(I[k].right+1);
			if(color[i]==0)
			{
				color[i]=I[k].color;
				if(u!=v)
				{
					next[u]=next[v];
					join(u,v);
				}
				++i;
			}
			else
			{
				i=next[u];
				if(u!=v)
				{
					next[u]=next[v];
					join(u,v);
				}
			}
		}
	}
}


int main()
{
	freopen("curcubeu.in","r",stdin);
	freopen("curcubeu.out","w",stdout);
	scanf("%d %d %d %d",&N,&A,&B,&C);
	init();
	solve();
	for(int i=1;i<N;i++)
		printf("%d\n",color[i]);
	return 0;
}