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: Accepted)

#include <stdio.h>
#include <stdlib.h>

int main(){
	int N, D1, D2, i, j, k, l, f, g;
	scanf("%d%d%d", &N, &D1, &D2);
	for(k = 0; (D1 & 1) == 0; D1 >>= 1, k++){}
	for(l = 0; (D2 & 1) == 0; D2 >>= 1, l++){}
	int num = 0;
	for(i = 0; i < 2 * N; i++){
		for(j = 0; j < 2 * N; j++){
			if(num == N * N){
				return 0;
			}
			if(((((k&1)|(((i>>(k>>1))^(j>>(k>>1)))&1))&(((k^1)&1)|((i>>(k>>1))&1)))|(((l&1)|(((i>>(l>>1))^(j>>(l>>1)))&1))&(((l^1)&1)|((i>>(l>>1))&1))))==0){
				printf("%d %d\n", i, j);
				num++;
			}
		}
	}
	return 0;
}