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<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int n,D1,D2;
int vis[605][605],vst[605][605],dx[4]={1,1,-1,-1},dy[4]={1,-1,1,-1},num[5][5];
void dfs(int x,int y,int D)
{
	for(int i=0;i*i<=D;i++)
	{
		int len=D-i*i;
		int j=sqrt(len);if(j*j!=len) continue;
		for(int k=0;k<4;k++)
		{
			int nx=x+i*dx[k],ny=y+j*dy[k];
			if(nx<1||nx>n||ny<1||ny>n) continue;
			if(!vis[nx][ny])
			{
				vis[nx][ny]=3-vis[x][y];
				dfs(nx,ny,D);
			}
		}
	}
}
int main()
{
	scanf("%d%d%d",&n,&D1,&D2);
	n<<=1;
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
	{
		if(!vis[i][j])
		{
			vis[i][j]=1;
			dfs(i,j,D1);
		}
	}
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++) vst[i][j]=vis[i][j];
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
	{
		if(!vis[i][j])
		{
			vis[i][j]=1;
			dfs(i,j,D2);
		}
	}
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++) num[vis[i][j]][vst[i][j]]++;
	bool flag=1;
	for(int a=1;a<=2;a++)
	for(int b=1;b<=2&&flag;b++)
	{
		if(num[a][b]<n*n/4) continue;
		int nums=0;
		for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		{
			if(nums==n*n/4) break;
			if(vis[i][j]==a&&vst[i][j]==b) printf("%d %d\n",i-1,j-1),nums++;
		}
		flag=0;
	}
	return 0;
}