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;

#define mp make_pair
#define fi first
#define se second
#define pb push_back

int dx[4] = {1, -1, -1, 1};
int dy[4] = {1, -1, 1, -1};
int n, d1, d2;
vector<pair<int, int> > has;
int vs[700][700];

int main() {
  cin >> n >> d1 >> d2;

  for (int i = 0; i < 400; i++) {
    for (int j = 0; j < 400; j++) {
      if (i * i + j * j == d1) {
        has.pb(mp(i, j));
      }
      if (i * i + j * j == d2) {
        has.pb(mp(i, j));
      }
    }
  }

  for (int i = 0; i < 2 * n; i++) {
    for (int j = 0; j < 2 * n; j++) {
      if (!vs[i][j]) {
        cout << i << ' ' << j << '\n';
        for (auto v : has) {
          for (int d = 0; d < 4; d++) {
            int ni = i + dx[d] * v.fi;
            int nj = j + dy[d] * v.se;
            if (ni >= 0 && nj >= 0 && ni < 2 * n && nj < 2 * n) {
              vs[ni][nj] = 1;
            }
          }
        }
      }
    }
  }

  return 0;
}