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 Ruby (2.3.3) (judge verdict: Wrong Answer)

def main()
  n, d1, d2 = gets.split.map(&:to_i)

  arr = Array.new(2*n){Array.new(2*n){1}}

  for y in 0..(2*n-1) do
    for x in 0..(2*n-1) do
      next if arr[y][x] == 0
      for i in 0..(2*n-1) do
        j1 = d1 - (x-i)**2
        if j1 >= 0
          j1 = Math.sqrt(j1)
          if (j1%1 == 0) && arr[y+j1]
            arr[y+j1][i] = 0
          end
        end

        j2 = d2 - (x-i)**2
        if j2 >= 0
          j2 = Math.sqrt(j2)
          if (j2%1 == 0) && arr[y+j2]

            arr[y+j2][i] = 0
          end
        end
      end
    end
  end

  c = 0
  for i in 0..(2*n-1) do
    for j in 0..(2*n-1) do
      if arr[j][i] == 1
        print i, " ", j, "\n"
        c += 1
        if c >= n**2
          return true
        end
      end
    end
  end
end

main()