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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N = 600 + 7;
const int M = 2e5 + 7;

bool bad[N][N];
int sq[M];

int main()
{
	ios::sync_with_stdio(false);

	memset(sq, -1, sizeof sq);
	for (int i=0; i*i<M; i++)
		sq[i*i] = i;
	int n, d[2];
	cin >> n >> d[0] >> d[1];
	int cnt = 0;
	for (int i=0; i<2*n; i++)
		for (int j=0; j<2*n; j++)
		{
			if (bad[i][j])
				continue;
			if (cnt < n * n)
				cout << i << " " << j << "\n";
			cnt++;
			for (int k=0; k<2*n; k++)
			{
				for (int l=0; l<2; l++)
				{
					int x = k * k;
					int y = d[l] - x;
					if (y < 0 || sq[y] == -1)
						continue;
					int o = sq[y];
					if (i + k < 2 * n)
					{
						if (j + o < 2 * n)
							bad[i+k][j+o] = true;
						if (j - o >= 0)
							bad[i+k][j+o] = true;
					}
				}
			}
		}
	return 0;
}