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<cstdio>
#include<vector>
#include<cmath>
#define pr pair<int,int>
using namespace std;
int n,d1,d2,cnt;
const int N=605;
int id[N][N],col[N*N][3];
vector<pr>d[2];
void dfs(int x,int opt)
{
	int X=(x-1)/(2*n)+1,Y=(x-1)%(2*n)+1,tox,toy;
	for(int i=0,siz=d[opt].size();i<siz;++i)
	{
		tox=X+d[opt][i].first;toy=Y+d[opt][i].second;
		if(1<=tox&&tox<=2*n&&1<=toy&&toy<=2*n&&!col[id[tox][toy]][opt])
			col[id[tox][toy]][opt]=3-col[x][opt],dfs(id[tox][toy],opt);
	}
}
int main()
{
    cin>>n>>d1>>d2;
    for(int i=0,j;i<=d1;++i)
    {
    	j=sqrt(d1-i*i);
    	if(i*i+j*j==d1)
    	{
    		d[1].push_back(pr(i,j));
    		d[1].push_back(pr(-i,j));
    		d[1].push_back(pr(i,-j));
    		d[1].push_back(pr(-i,-j));
		}
	}
	for(int i=0,j;i<=d2;++i)
    {
    	j=sqrt(d2-i*i);
    	if(i*i+j*j==d2)
    	{
    		d[2].push_back(pr(i,j));
    		d[2].push_back(pr(-i,j));
    		d[2].push_back(pr(i,-j));
    		d[2].push_back(pr(-i,-j));
		}
	}
	for(int i=1;i<=2*n;++i)
		for(int j=1;j<=2*n;++j)
			id[i][j]=(i-1)*2*n+j;
	for(int i=1;i<=2*n;++i)
		for(int j=1;j<=2*n;++j)
		{
			if(!col[id[i][j]][1])col[id[i][j]][1]=1,dfs(id[i][j],1);
			if(!col[id[i][j]][2])col[id[i][j]][2]=1,dfs(id[i][j],2);
		}
    for(int i=1;i<=2*n;++i)
    	for(int j=1;j<=2*n;++j)
    	{
    		if(col[id[i][j]][1]==1&&col[id[i][j]][2]==1)printf("%d %d\n",i-1,j-1),++cnt;
    		if(cnt==n*n)return 0;
		}
    return 0;
}