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: Accepted)

#include<bits/stdc++.h>

using namespace std;

int nr, N, D1, D2, cod[609][609], x[609 * 609], y[609 * 609], c[2][609 * 609], sqr[609 * 609];
vector < int > v[2][609 * 609], h[4];

void add (int line, int i, int j)
{
    v[line][i].push_back (j);
    v[line][j].push_back (i);
}

void dfs (int lin, int nod, int col)
{
    if (c[lin][nod] != 0)
        return ;
    c[lin][nod] = col;
    for (auto it : v[lin][nod])
        dfs (lin, it, 3 - col);
}

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d %d %d", &N, &D1, &D2);
for (int i=0; i<2 * N; i++)
    sqr[i * i] = i + 1;
vector < pair < int, int > > dlt[2];
for (int dx=0; dx<2 * N; dx++)
{
    if (dx * dx <= D1 && sqr[D1 - dx * dx] > 0)
    {
        int dy = sqr[D1 - dx * dx] - 1;
        dlt[0].push_back ({dx, dy});
    }
    if (dx * dx <= D2 && sqr[D2 - dx * dx] > 0)
    {
        int dy = sqr[D2 - dx * dx] - 1;
        dlt[1].push_back ({dx, dy});
    }
}
for (int i=0; i<2 * N; i++)
    for (int j=0; j<2 * N; j++)
        cod[i][j] = ++nr,
        x[nr] = i, y[nr] = j;
for (int i=0; i<2 * N; i++)
    for (int j=0; j<2 * N; j++)
    {
        for (int k=0; k<2; k++)
            for (auto d : dlt[k])
            {
                if (i >= d.first && j >= d.second)
                    add (k, cod[i][j], cod[i - d.first][j - d.second]);
                if (i >= d.first && j + d.second < 2 * N)
                    add (k, cod[i][j], cod[i - d.first][j + d.second]);
            }
    }
for (int i=0; i<2; i++)
    for (int j=1; j<=nr; j++)
        if (c[i][j] == 0)
            dfs (i, j, 1);
for (int i=1; i<=nr; i++)
    h[(c[0][i] - 1) * 2 + c[1][i] - 1].push_back (i);
int lin = -1;
for (int i=0; i<4; i++)
    if (h[i].size () >= N * N)
        lin = i;
int sz = 0;
for (auto it : h[lin])
    if (++sz <= N * N)
        printf ("%d %d\n", x[it], y[it]);
return 0;
}