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++ (GCC 5.4.1) (judge verdict: Runtime Error)

#include <cstdio>
#include <cstring>
#include <cassert>
#include <algorithm>
#include <vector>
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define sz(a) ((int) (a).size())

const int N = 300;
int used[2 * N][2 * N], bad[2 * N], cnt[2 * N];



int main() {
	int n, d1, d2;
	scanf("%d%d%d", &n, &d1, &d2);
	if ((d1 % 4 != 0) && (d2 % 4 != 0)) {
		for (int i = 0; i < 2 * n; ++i) {
			for (int j = 0; j < 2 * n; ++j) {
				if ((i % 2 == 0) && (j % 2 == 0)) {
					printf("%d %d\n", i, j);
				}
			}
		}
		return 0;
	}
	if (d2 % 4 != 0) {
		swap(d1, d2);
	}
	vector< pair<int, int> > a, dirs, output;
	for (int x = 0; x < 2 * n; ++x) {
		for (int y = 0; y < 2 * n; ++y) {
			int d = x * x + y * y;
			if ((d == d1) || (d == d2)) {
				dirs.pb(mp(x, y));
			}
		}
	}
	for (int i = 0; i < 2 * n; ++i) {
		for (int j = 0; j < 2 * n; ++j) {
			if (((i % 2 == 0) && (j % 2 == 0)) || (d2 == 1)) {
				a.pb(mp(i, j));
			}
		}
	}
	int max_c = n * n;
		for (int i = 0; i < sz(a); ++i) {
			int x = a[i].fs, y = a[i].sc;
			if (used[x][y]) continue;
			if (max_c > 0) {printf("%d %d\n", x, y), --max_c;}
			if (max_c > 0) {printf("%d %d\n", x + 1, y); --max_c;}
			for (int j = 0; j < sz(dirs); ++j) {
				for (int sx = -1; sx <= 1; sx += 2) {
					for (int sy = -1; sy <= 1; sy += 2) {
						int nx = x + sx * dirs[j].fs, ny = y + sy * dirs[j].sc;
						if ((nx >= 0) && (nx < 2 * n) && (ny >= 0) && (ny < 2 * n)) {
							used[nx][ny] = 1;
						}
					}
				}
			}
		}
	assert(max_c == 0);
	return 0;
}