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;

typedef pair<int,int> pii;
const int mx = 605;

int mark[mx][mx];

vector<pii> adj[2];
vector<pii> P[4];

int D[2];
int n;

void init(){
    cin >> n >> D[0] >> D[1];
    for(int i = 0; i < 2 * n; i++) fill(mark[i],mark[i] + 2*n, -1);

    for(int b = 0; b < 2; b++){
        for(int i = (int)(sqrt(D[b]+1e-9)); i >= 0; i--){
            for(int j = i; j >= 0; j--){
                if(i * i + j * j == D[b]){
                    adj[b].emplace_back(i,j);
                    adj[b].emplace_back(j,i);
                    adj[b].emplace_back(-i,j);
                    adj[b].emplace_back(j,-i);
                    adj[b].emplace_back(i,-j);
                    adj[b].emplace_back(-j,i);
                    adj[b].emplace_back(-i,-j);
                    adj[b].emplace_back(-j,-i);

                    break;
                }
            }
        }

        sort(adj[b].begin(),adj[b].end());
        adj[b].erase(unique(adj[b].begin(),adj[b].end()),adj[b].end());
/*
        for(pii &p : adj[b]){
            cout << p.first << ' ' << p.second << '\n';
        }
        cout << "-----------\n";
*/
    }
}

void dfs(int r, int c, int ty, int v, int tog, int bv){
    mark[r][c] = v;
    P[v].emplace_back(r,c);
    int nr, nc;
    for(pii &p : adj[ty]){
        nr = r + p.first;
        nc = c + p.second;
        if(nr < 0 || nr >= 2 * n || nc < 0 || nc >= 2 * n) continue;
        if(mark[nr][nc] != bv) continue;
        dfs(nr,nc,ty,tog,v,bv);
    }
}

void pro(){
    for(int i = 0; i < 2*n; i++){
        for(int j = 0; j < 2*n; j++){
            if(mark[i][j] != -1) continue;
            dfs(i,j,0,0,1,-1);
        }
    }
    int dom = (P[1].size() >= P[0].size() ? 1 : 0);
    for(pii &p : P[dom]){
        if(mark[p.first][p.second] != dom) continue;
        dfs(p.first,p.second,1,2,3,1);
    }
    int ddom = (P[3].size() >= P[2].size() ? 3 : 2);

    assert(P[ddom].size() >= n * n);
    int C = n * n;
    for(int i = 0; i < C; i++){
        cout << P[ddom][i].first << ' ' << P[ddom][i].second << '\n';
    }
}

int main(){
    ios_base::sync_with_stdio(false);cin.tie(0);
    init();
    pro();
}