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.

Examples

Input

2 1 2


Output

0 0
0 2
2 0
2 2


Input

3 1 5


Output

0 0
0 2
0 4
1 1
1 3
1 5
2 0
2 2
2 4
#include<cstdio>
#include<cstdlib>
#include<cmath>
using namespace std;
int n,m,d1,d2,col[610][610],col2[610][610];
int cnt,dx[1100000],dy[1100000];
void dfs(int x,int y){
	int nx,ny;
	for (int i=1;i<=cnt;i++){
		nx=x+dx[i]; ny=y+dy[i];
		if (nx>=0&&nx<m&&ny>=0&&ny<m&&col[nx][ny]==-1){
			col[nx][ny]=col[x][y]^1;
			dfs(nx,ny);
		}
	}
}
void work(int d){
	cnt=0;
	for (int i=0;i<=d;i++){
		int j=sqrt(d-i*i);
		if (i*i+j*j==d){
			cnt++; dx[cnt]=i; dy[cnt]=j;
			cnt++; dx[cnt]=-i; dy[cnt]=j;
			cnt++; dx[cnt]=i; dy[cnt]=-j;
			cnt++; dx[cnt]=-i; dy[cnt]=-j;
		}
	}
	for (int i=0;i<m;i++)
		for (int j=0;j<m;j++)
			col[i][j]=-1;
	for (int i=0;i<m;i++)
		for (int j=0;j<m;j++)
			if (col[i][j]==-1){
				col[i][j]=0;
				dfs(i,j);
			}
}
void print(){
	int tot=0;
	for (int i=0;i<m;i++)
		for (int j=0;j<m;j++)
			if (!col[i][j]&&!col2[i][j]){
				printf("%d %d\n",i,j);
				tot++; if (tot==n*n) return;
			}
}
int main(){
	scanf("%d%d%d",&n,&d1,&d2); m=n*2;
	work(d1);
	for (int i=0;i<m;i++)
		for (int j=0;j<m;j++)
			col2[i][j]=col[i][j];
	work(d2);
	print();
	return 0;
}