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<vector>
using namespace std;
int n;
int color[2][1000][1000];
int db[2<<17];
int dx[]={1,1,-1,-1},dy[]={1,-1,1,-1};
void dfs(int x,int y,int d,int col,int c)
{
	color[c][x][y]=col;
	for(int i=0;i*i<=d;i++)
	{
		if(db[d-i*i]<0)continue;
		int j=db[d-i*i];
		for(int r=0;r<4;r++)
		{
			int tx=x+dx[r]*i,ty=y+dy[r]*j;
			if(tx<0||ty<0||tx>=2*n||ty>=2*n||color[c][tx][ty])continue;
			dfs(tx,ty,d,col==1?2:1,c);
		}
	}
}
int main()
{
	cin>>n;
	for(int i=0;i<2<<17;i++)db[i]=-1;
	for(int i=0;i*i<2<<17;i++)db[i*i]=i;
	for(int i=0;i<2;i++)
	{
		int d;cin>>d;
		for(int x=0;x<2*n;x++)for(int y=0;y<2*n;y++)if(!color[i][x][y])dfs(x,y,d,1,i);
	}
	for(int i=1;i<=2;i++)for(int j=1;j<=2;j++)
	{
		vector<pair<int,int> >a;
		for(int x=0;x<2*n;x++)
		{
			for(int y=0;y<2*n;y++)
			{
				if(color[0][x][y]==i&&color[1][x][y]==j)a.push_back(make_pair(x,y));
				if(a.size()==n*n)
				{
					for(int i=0;i<n*n;i++)cout<<a[i].first<<" "<<a[i].second<<endl;
					return 0;
				}
			}
		}
	}
}
