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>
#include <complex>
#define pb push_back
#define pll pair <ll, ll>
#define MOMI using namespace std;
#define mp make_pair
#define pyshnapyshnakaa ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
#pragma optimize("TKACHENKO-GORYACHENKO")
#pragma GCC optimize("O3")

//#pragma GCC optimize("unroll-loops")
//#define double long double

typedef long long ll;

typedef long double ld;

using namespace std;

const ll maxn = 700;
const ll inf = 1e16;

ll n, m, k;

ll d1, d2;

bool used[maxn][maxn];

vector <pll> ANS;

vector <pll> X;

ll M[maxn][maxn];

bool in(ll x, ll y){
    return x >= 0 && x < 2 * n && y >= 0 && y < 2 * n;
}

bool cmp(pll a, pll b){
    return M[a.first][a.second] > M[b.first][b.second];
}

int main() {
    ll q, w, e, a, b, t;
    cin >> n >> d1 >> d2;
    vector <pll> C;
    for (q = 0; q * q <= d1; q++){
        for (w = 0; w * w <= d1; w++){
            if (q * q + w * w == d1){
                C.pb(mp(q, w));
            }
        }
    }
    for (q = 0; q * q <= d2; q++){
        for (w = 0; w * w <= d2; w++){
            if (q * q + w * w == d2){
                C.pb(mp(q, w));
            }
        }
    }
    for (q = 0; q < 2 * n; q++){
        for (w = 0; w < 2 * n; w++){
            pll p = mp(q, w);
            M[q][w] = 0;
            for (e = 0; e < C.size(); e++){
                ll x = q + C[e].first, y = w + C[e].second;
                if (!in(x, y)){
                    M[q][w]++;
                }
                x = q - C[e].first, y = w - C[e].second;
                if (!in(x, y)){
                    M[q][w]++;
                }
                x = q + C[e].first, y = w - C[e].second;
                if (!in(x, y)){
                    M[q][w]++;
                }
                x = q - C[e].first, y = w + C[e].second;
                if (!in(x, y)){
                    M[q][w]++;
                }
            }
            X.pb(mp(q, w));
        }
    }
    sort(X.begin(), X.end(), cmp);
    for (t = 0; t < X.size(); t++){
        ll q = X[t].first, w = X[t].second;
        if (!used[q][w]){
            ANS.pb(mp(q, w));
            for (e = 0; e < C.size(); e++){
                ll x = q + C[e].first, y = w + C[e].second;
                if (in(x, y)){
                    used[x][y] = 1;
                }
                x = q - C[e].first, y = w - C[e].second;
                if (in(x, y)){
                    used[x][y] = 1;
                }
                x = q + C[e].first, y = w - C[e].second;
                if (in(x, y)){
                    used[x][y] = 1;
                }
                x = q - C[e].first, y = w + C[e].second;
                if (in(x, y)){
                    used[x][y] = 1;
                }
            }
        }
    }
    if (ANS.size() < n * n){
        return -1;
    }
    for (q = 0; q < n * n; q++){
        cout << ANS[q].first << " " << ANS[q].second << endl;
    }
//    pyshnapyshnakaa;
    return 0;
}