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: Wrong Answer)

#include<stdio.h>

int main() {
	int n, d1, d2,max;
	int dx1, dy1, dx2, dy2;
	int s[1500][1500] = {};
	scanf("%d%d%d", &n, &d1, &d2);

	for (int i = 0; i < 2 * n; i++) {
		for (int j = i; j < 2 * n; j++) {
			if (d1 == i * i + j * j) {
				dx1 = i;
				dy1 = j;
				goto next;
			}
		}
	}

next:

	for (int i = 0; i < 2 * n; i++) {
		for (int j = i; j < 2 * n; j++) {
			if (d2 == i * i + j * j) {
				dx2 = i;
				dy2 = j;
				goto next2;
			}
		}
	}
next2:

	for (int i = 0; i < 2 * n; i++) {
		for (int j = 0; j < 2 * n; j++) {
			if (s[j][i] == 0) {
				s[j][i] = 0;
				s[j + dx1][i + dy1] = 2;
				s[j + dy1][i + dx1] = 2;
				if (j - dx1 >= 0) {
					s[j - dx1][i + dy1] = 2;
				}
				if (j - dy1 >= 0) {
					s[j - dy1][i + dx1] = 2;
				}
				if (i - dx1 >= 0) {
					s[j + dy1][i - dx1] = 2;
				}
				if (i - dy1 >= 0) {
					s[j + dx1][i - dy1] = 2;
				}
				if (i - dx1 >= 0&&j-dy1>=0) {
					s[j - dy1][i - dx1] = 2;
				}
				if (i - dy1 >= 0 && j - dx1 >= 0) {
					s[j + dx1][i - dy1] = 2;
				}
				s[j + dx2][i + dy2] = 2;
				s[j + dy2][i + dx2] = 2;
				if (j - dx2 >= 0) {
					s[j - dx2][i + dy2] = 2;
				}
				if (j - dy2 >= 0) {
					s[j - dy2][i + dx2] = 2;
				}
				if (i - dx2 >= 0) {
					s[j + dy2][i - dx2] = 2;
				}
				if (i - dy2 >= 0) {
					s[j + dx2][i - dy2] = 2;
				}
				if (i - dx2 >= 0 && j - dy2 >= 0) {
					s[j - dy2][i - dx2] = 2;
				}
				if (i - dy2 >= 0 && j - dx2 >= 0) {
					s[j + dx2][i - dy2] = 2;
				}
			}
		}
	}

	max = n*n;
	for (int i = 0; i < 2 * n; i++) {
		for (int j = 0; j < 2 * n; j++) {
			if (s[j][i] == 0) {
				printf("%d %d\n", j, i);
				max--;
				if (max == 0) { return 0; }
			}
		}
	}
}