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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
#define Maxn 607
using namespace std;
int n,d1,d2;
int flag1[Maxn][Maxn],flag2[Maxn][Maxn];
void dfs1(int x,int y,int now)
{
	if (flag1[x][y]!=-1) return;
	flag1[x][y]=now;
	for (int j=0;j<2*n;j++)
	{
		long long del=1LL*d1-1LL*(x-j)*(x-j);
		if (del<0) continue;
		if (1LL*(int)sqrt(del)*(int)sqrt(del)!=del) continue;
		del=(int)sqrt(del);
		if (y-del>=0) dfs1(j,y-del,now^1);
		if (y+del<2*n) dfs1(j,y+del,now^1);
	}
}
void dfs2(int x,int y,int now)
{
	if (flag2[x][y]!=-1) return;
	flag2[x][y]=now;
	for (int j=0;j<2*n;j++)
	{
		long long del=1LL*d2-1LL*(x-j)*(x-j);
		if (del<0) continue;
		if (1LL*(int)sqrt(del)*(int)sqrt(del)!=del) continue;
		del=(int)sqrt(del);
		if (y-del>=0) dfs2(j,y-del,now^1);
		if (y+del<2*n) dfs2(j,y+del,now^1);
	}
}
int main()
{
	scanf("%d%d%d",&n,&d1,&d2);
	memset(flag1,-1,sizeof(flag1));
	memset(flag2,-1,sizeof(flag2));
	for (int i=0;i<2*n;i++)
		for (int j=0;j<2*n;j++)
		{
			if (flag1[i][j]==-1) dfs1(i,j,0);
			if (flag2[i][j]==-1) dfs2(i,j,0);
		}
	int res1,res2;
	int cnt1=0,cnt2=0,cnt3=0,cnt4=0;
	for (int i=0;i<2*n;i++)
		for (int j=0;j<2*n;j++)
		{
			if (flag1[i][j]) ++cnt1; else ++cnt2;
			if (flag2[i][j]) ++cnt3; else ++cnt4;
		}
	if (cnt1>=n*n&&cnt3>=n*n) res1=1,res2=1;
	if (cnt1>=n*n&&cnt4>=n*n) res1=1,res2=0;
	if (cnt2>=n*n&&cnt3>=n*n) res1=0,res2=1;
	if (cnt2>=n*n&&cnt4>=n*n) res1=0,res2=0;
	int ans=0;
	for (int i=0;i<2*n;i++)
		for (int j=0;j<2*n;j++)
			if (flag1[i][j]==res1&&flag2[i][j]==res2) 
			{
				printf("%d %d\n",i,j);
				++ans;
				if (ans==n*n) return 0;
			}
	return 0;
}