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<algorithm>
#include<cstring>
using namespace std;
typedef pair<int,int> P;
P q[101000];int n;
int col[620][620],s[620][620],top=0;
void dfs(int x,int y,int t)
{
	col[x][y]=t;
	for(int i=1;i<=top;i++)
	{
		int dx=x+q[i].first,dy=y+q[i].second;
		if(dx<0||dx>=2*n||dy<0||dy>=2*n||col[dx][dy]!=-1)continue;
		dfs(dx,dy,t^1);
	}
}
void solve(int d)
{ 
	top=0;
	for(int i=0;i*i<=d;i++)
	for(int j=0;j*j<=d;j++)
		if(i*i+j*j==d)
		top++,q[top].first=i,q[top].second=j,top++,q[top].first=-i,q[top].second=-j,
		top++,q[top].first=-i,q[top].second=j,top++,q[top].first=i,q[top].second=-j,
	memset(col,-1,sizeof(col));
	for(int i=0;i<2*n;i++)
	for(int j=0;j<2*n;j++)if(col[i][j]==-1)dfs(i,j,1);
}
int main()
{
	int d1,d2;scanf("%d%d%d",&n,&d1,&d2); 
	solve(d1);
	for(int i=0;i<2*n;i++)
	for(int j=0;j<2*n;j++)s[i][j]=col[i][j];
	solve(d2);
	for(int i=0;i<2*n;i++)
	for(int j=0;j<2*n;j++)s[i][j]=s[i][j]*2+col[i][j]; 
	for(int k=0;k<=3;k++)
	{
		int sum=0;
		for(int i=0;i<2*n;i++)
		for(int j=0;j<2*n;j++)
			if(s[i][j]==k)sum++;
		if(sum>=n*n)
		{
			sum=n*n;
			for(int i=0;i<2*n;i++)
			for(int j=0;j<2*n;j++)
				if(s[i][j]==k&&sum>0)sum--,printf("%d %d\n",i,j);
			return 0;
		 } 
	}
	return 0;
}