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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

typedef long long lli;
typedef double lld;
typedef vector<lli> vll;
typedef vector<bool> vbl;
typedef vector<double> vdl;
typedef vector<vector<lli>> mat;
typedef vector<vdl> mad;
typedef unordered_map<lli,unordered_map<lli,lli>> graph;
typedef complex<double> cmp;
typedef vector<cmp> vcl;

lli n,d1,d2;
mat c;
graph g;
vll argidx;
mat f;

int main(){
  cin >> n >> d1 >> d2;
  lli d = d1;
  for(lli x = 0,y = d;x <= d;x++){
    while(x*x+y*y > d && y >= 0) y--;
    if(y < 0) break;
    if(x*x+y*y == d){
      c.push_back(vll({x,y}));
      c.push_back(vll({-x,y}));
      c.push_back(vll({x,-y}));
      c.push_back(vll({-x,-y}));
    }
  }
  d = d2;
  for(lli x = 0,y = d;x <= d;x++){
    while(x*x+y*y > d && y >= 0) y--;
    if(y < 0) break;
    if(x*x+y*y == d){
      c.push_back(vll({x,y}));
      c.push_back(vll({-x,y}));
      c.push_back(vll({x,-y}));
      c.push_back(vll({-x,-y}));
    }
  }
  for(lli i = 0;i < 2*n;i++){
    for(lli j = 0;j < 2*n;j++){
      for(vll d : c){
        if(i+d[0] >= 0 && i+d[0] < 2*n && j+d[1] >= 0 && j+d[1] < 2*n) g[i*n*2+j][(i+d[0])*n*2+j+d[1]] = 1;
      }
    }
  }
  argidx = vll(2*2*n*n);
  iota(argidx.begin(),argidx.end(), 0ll);
  sort(argidx.begin(),argidx.end(),[](lli a, lli b){
    return g[a].size() < g[b].size();
  });
  f = mat(2*n,vll(2*n));
  for(lli i = 0,j = 0;j < n*n;j++){
    while(i < n*n*2*2 && f[argidx[i]/(n*2)][argidx[i]%(n*2)] != 0) i++;
    if(i == n*n*2*2) exit(1);
    lli x = argidx[i]/(n*2);
    lli y = argidx[i]%(n*2);
    f[x][y] = 1;
    for(auto next : g[argidx[i]]){
      f[next.first/(n*2)][next.first%(n*2)] = 1;
    }
    cout << x << " " << y << endl;
  }
}