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: Runtime Error)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 350
int map[N][N],n;
void work(int D) {
	while(D%4==0) D/=4;
	int i,j;
	if(D&1) {
		for(i=0;i<2*n;i++) {
			for(j=0;j<2*n;j++) {
				if((i+j)&1) map[i][j]=1;
			}
		}
	}else {
		for(i=0;i<2*n;i++) {
			for(j=0;j<2*n;j++) {
				if(i&1) map[i][j]=1;
			}
		}
	}
}
int main() {
	int d1,d2;
	scanf("%d%d%d",&n,&d1,&d2);
	work(d1); work(d2);
	int i,j,sum=0;
	for(i=0;i<2*n;i++) {
		for(j=0;j<2*n;j++) {
			if(!map[i][j]) {
				printf("%d %d\n",i,j); sum++;
				if(sum==n*n) return 0;
			}
		}
	}
}