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>
#define debug(...) fprintf(stderr, __VA_ARGS__)

using namespace std;
const int N = 605;

inline int input() {
	int x; char ch; while (!isdigit(ch = getchar()));
	for (x = ch ^ 48; isdigit(ch = getchar()); x = x * 10 + (ch ^ 48));
	return x;
}

inline void chkmin(int &x, int y) { x = x < y ? x : y; }
template <class T> T sqr(T x) { return x * x; }

int n, m, D1, D2;
int ansx[N * N], ansy[N * N], tot;
bool vis[N][N];

void Add(int x, int y) {
	if (tot == m * m) return void();
	ansx[++tot] = x, ansy[tot] = y, vis[x][y] = true;
	for (int i = 0, Dy; i <= n; ++i) {
		Dy = D1 - sqr(abs(x - i));
		if (sqr((int)sqrt(Dy)) == Dy) {
			Dy = sqrt(Dy);
			if (y + Dy <= n) vis[i][y + Dy] = true;
			if (y - Dy >= 0) vis[i][y - Dy] = true;
		}
		Dy = D2 - sqr(abs(x - i));
		if (sqr((int)sqrt(Dy)) == Dy) {
			Dy = sqrt(Dy);
			if (y + Dy <= n) vis[i][y + Dy] = true;
			if (y - Dy >= 0) vis[i][y - Dy] = true;
		}
	}
}

int main() {
#ifdef local
	freopen("in", "r", stdin);
	freopen("out", "w", stdout);
//	freopen("log", "w", stderr);
#endif
	ios::sync_with_stdio(false);

	cin >> n >> D1 >> D2;
	m = n, n = n * 2 - 1;

	for (int i = 0; i < m; ++i) {
		if (!vis[i][i]) Add(i, i);
		if (!vis[i][n - i]) Add(i, n - i);
		if (!vis[n - i][i]) Add(n - i, i);
		if (!vis[n - i][n - i]) Add(n - i, n - i);

		for (int x = i + 1; x < n - i; ++x) {
			if (!vis[x][i]) Add(x, i);
			if (!vis[x][n - i]) Add(x, n - i);
		}
		for (int y = i + 1; y < n - i; ++y) {
			if (!vis[i][y]) Add(i, y);
			if (!vis[n - i][y]) Add(n - i, y);
		}
	}

	for (int i = 1; i <= tot; ++i)
		printf("%d %d\n", ansx[i], ansy[i]);
	return 0;
}