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 long long ll;

const int N = 1000;

int kek[N][N];
int n, d1, d2;
vector<pair<int, int>> v;
vector<pair<int, int>> ans;

void dfs(int x, int y) {
    for (auto p : v) {
        if (x + p.first < 0 || y + p.second < 0 || x + p.first >= 2 * n || y + p.second >= 2 * n || kek[x + p.first][y + p.second] != 0) {
            continue;
        }
        kek[x + p.first][y + p.second] = -kek[x][y];
        if (kek[x][y] == -1 && ans.size() < n * n) {
            ans.push_back({x + p.first, y + p.second});
        }
        dfs(x + p.first, y + p.second);
    }
}

int main() {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> d1 >> d2;
    for (int i = 0; i * i <= d1; ++i) {
        int rt = sqrt(d1 - i * i);
        if (rt * rt + i * i == d1 && max(i, rt) < 2 * n) {
            v.push_back({i, rt});
            v.push_back({-i, rt});
            v.push_back({i, -rt});
            v.push_back({-i, -rt});
        }
    }
    for (int i = 0; i * i <= d2; ++i) {
        int rt = sqrt(d2 - i * i);
        if (rt * rt + i * i == d2 && max(i, rt) < 2 * n) {
            v.push_back({i, rt});
            v.push_back({-i, rt});
            v.push_back({i, -rt});
            v.push_back({-i, -rt});
        }
    }
    for (int i = 0; i < n * 2; ++i) {
        for (int j = 0; j < n * 2; ++j) {
            if (ans.size() >= n * n) {
                break;
            }
            if (kek[i][j] == 0) {
                kek[i][j] = 1;
                ans.push_back({i, j});
                dfs(i, j);
            }
        }
    }
    if (ans.size() < n * n) {
        cout << 2 / 0;
    } else {
        for (int i = 0; i < n * n; ++i) {
            cout << ans[i].first << ' ' << ans[i].second << '\n';
        }
    }
    return 0;
}