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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
const double EPS=1e-7;
 
int main()
{
  int n,d1,d2;
  scanf("%d%d%d",&n,&d1,&d2);
  if(d1%4!=0&&d2%4!=0){
    for(int i=0;i<n;i++){
      for(int j=0;j<n;j++){
	printf("%d %d\n",2*i,2*j);
      }
    }
    return 0;
  }
  vector<pair<int,int> > V;
  for(int t=0;t<2;t++){
    int dt=(t==0?d1:d2);
    for(int i=0;i*i<=dt;i++){
      int d=dt-i*i,j=(int)(sqrt((double)d)+EPS);
      if(j*j!=d)continue;
      V.push_back(make_pair(i,j));
      if(i>0){
	V.push_back(make_pair(-i,j));
	if(j>0){
	  V.push_back(make_pair(-i,-j));
	}
      }
      if(j>0){
	V.push_back(make_pair(i,-j));
      }
    }
  }
  while(1){
    vector<int> R(4*n*n);
    for(int i=0;i<4*n*n;i++)R[i]=i;
    random_shuffle(R.begin(),R.end());
    vector<pair<int,int> > ans;
    vector<vector<bool> > B(2*n,vector<bool>(2*n,0));
    for(int D=0;D<4*n*n&&ans.size()<n*n;D++){
      int i=R[D]/(2*n),j=R[D]%(2*n);
      if(!B[i][j]){
	ans.push_back(make_pair(i,j));
	for(int t=0;t<V.size();t++){
	  int k=i+V[t].first,l=j+V[t].second;
	  if(0<=k&&k<2*n&&0<=l&&l<2*n){
	    B[k][l]=1;
	  }
	}
      }
    }
    if(ans.size()!=n*n)continue;
    for(int t=0;t<n*n;t++){
      printf("%d %d\n",ans[t].first,ans[t].second);
    }
    return 0;
  }
  return 0;
}