Problem:
Score : 800 points

Problem Statement

Takahashi is doing a research on sets of points in a plane.
Takahashi thinks a set S of points in a coordinate plane is a good set when S satisfies both of the following conditions:

The distance between any two points in S is not \sqrt{D_1}.

The distance between any two points in S is not \sqrt{D_2}.

Here, D_1 and D_2 are positive integer constants that Takahashi specified.

Let X be a set of points (i,j) on a coordinate plane where i and j are integers and satisfy 0 ≤ i,j < 2N.

Takahashi has proved that, for any choice of D_1 and D_2, there exists a way to choose N^2 points from X so that the chosen points form a good set.
However, he does not know the specific way to choose such points to form a good set.
Find a subset of X whose size is N^2 that forms a good set.

Constraints

1 ≤ N ≤ 300

1 ≤ D_1 ≤ 2×10^5

1 ≤ D_2 ≤ 2×10^5

All values in the input are integers.

Input

Input is given from Standard Input in the following format:

N D_1 D_2

Output

Print N^2 distinct points that satisfy the condition in the following format:

x_1 y_1
x_2 y_2
:
x_{N^2} y_{N^2}

Here, (x_i,y_i) represents the i-th chosen point.
0 ≤ x_i,y_i < 2N must hold, and they must be integers.
The chosen points may be printed in any order.
In case there are multiple possible solutions, you can output any.

Sample Input 1

2 1 2

Sample Output 1

0 0
0 2
2 0
2 2

Among these points, the distance between 2 points is either 2 or 2\sqrt{2}, thus it satisfies the condition.

Sample Input 2

3 1 5

Sample Output 2

0 0
0 2
0 4
1 1
1 3
1 5
2 0
2 2
2 4

Solution in C++ (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define pb push_back
#define p(x,y) ((x-1)*n+y)
using namespace std;
typedef double db;
const int P=1010,N=P*P,U=2e7+10;
int n,tot=0,B,las;
int pl[N],del[N];
struct Data
{
	vector<int> s[2];
}f[N];
struct MSolve
{
	int D,top,cnt;
	int cx[50],cy[50],col[N],vis[N];
	void dfs(int x,int y)
	{
		int id=p(x,y),i,v,t1,t2,nx,ny;
		vis[id]=1;
		//printf("P%d %d  %d\n",x,y,col[id]);
		f[tot].s[col[id]].pb(id);
		for(i=1;i<=top;i++)
		{
			for(t1=0;t1<2;t1++)
			{	
				for(t2=0;t2<2;t2++)
				{
					nx=x+(t1==0?cx[i]:-cx[i]);ny=y+(t2==0?cy[i]:-cy[i]);
					if(nx>=1&&nx<=n&&ny>=1&&ny<=n) 
					{
						v=p(nx,ny);col[v]=col[id]^1;
						if(!vis[v]) dfs(nx,ny);
					}
				}
			}
		}
	}
	void solve()
	{
		int i,j,w;
		cin>>D;
		if(D==las) return ;
		las=D;
      	
		for(i=0;i<n&&i*i<=D;i++)
		{
			w=floor(sqrt(D-i*i)+0.5);
			if(i*i+w*w==D) 
			{
				top++;cx[top]=i;cy[top]=w;
			}
		}
		
        tot++;f[tot].s[0].clear();f[tot].s[1].clear();
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				if(vis[p(i,j)]) continue;
				dfs(i,j);
			}
		}
       
		//en=clock();
		//printf("time%d\n",(int)(en-st));
	}
}h1,h2;
bool cmp(const int &a,const int &b){return f[a].s[0].size()+f[a].s[1].size()>f[b].s[0].size()+f[b].s[1].size();}
void dfs(int xh,int nas)
{
  //printf("tot%d\n",tot);
  
  //puts("OK");exit(0);
	if(nas<B) return ;
	if(xh==tot+1)
	{
         
		int num=0;
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				if(del[p(i,j)]==0) num++,printf("%d %d\n",i-1,j-1);
				if(num==B) exit(0);
			}
		}
		return ;
	}
	int val,i,j,id,vsi,x;
	for(i=0;i<2;i++)
	{
		id=pl[xh];
      
        vsi=f[id].s[i].size();
		val=nas;
		for(j=0;j<vsi;j++)
		{
          	x=f[id].s[i][j];
			if(!del[x]) val--;
			del[x]++;
		}
        
		dfs(xh+1,val);
		for(j=0;j<vsi;j++)
		{
          	x=f[id].s[i][j];
			del[x]--;
         	if(!del[x]) val++;
          	
		}
	}
}
int main()
{

	int i;
	cin>>n;n*=2;
	B=(n>>1)*(n>>1);
	h1.solve();h2.solve();
	for(i=1;i<=tot;i++) pl[i]=i;
	sort(pl+1,pl+tot+1,cmp);
   
	dfs(1,n*n);

	return 0;
}