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: Wrong Answer)

# include <iostream>
# include <cstdio>
# include <cmath>
using namespace std;
const int N = 2012;
const int M = 3e5 + 12;
bool map[N][N],vis[N][N];
int n,dt,d1,d2;
long long pw[M];
pair<int,int>que[M];
void mark(int x,int y)
{
	for(int i = 0;i < dt;i++)if(x + que[i].first < 2 * n && y + que[i].second < 2 * n)
	vis[x + que[i].first][y + que[i].second] = true;
}
int main()
{
    for(long long i = 0;i < M;i++)pw[i] = i * i;
	scanf("%d %d %d",&n,&d1,&d2);
	for(int i = 0,j;pw[i] <= pw[d1];i++)
	{
		j = sqrt(pw[d1] - pw[i]);
		if(pw[i] + pw[j] == pw[d1])que[dt++] = make_pair(i,j);
	}
	for(int i = 0,j;pw[i] <= pw[d2];i++)
	{
		j = sqrt(pw[d2] - pw[i]);
		if(pw[i] + pw[j] == pw[d2])que[dt++] = make_pair(i,j);
	}
	for(int i = 0;i < 2 * n;i++)
	  for(int j = 0;j < 2 * n;j++)if(!vis[i][j])
	  {
	  	map[i][j] = true; 
	  	mark(i,j);
	  }
	int dt = 0;
	for(int i = 0;i < 2 * n && dt < n * n;i++)
	 for(int j = 0;j < 2 * n && dt < n * n;j++)
	 if(map[i][j])printf("%d %d\n",i,j),dt++;
}