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<bits/stdc++.h>
using namespace std;
const int N=620;
int n,D1,D2;
int t1[N][N],t2[N][N];
void Solve(int D,int t[N][N])
{
	int i,j,k,K,len,x,y;
	K=0;
	while(D%2==0) K++,D>>=1;
	if(K&1)
	{
		len=1<<((K+1)>>1);
		for(i=0;i<2*n;i+=len)
		{
			for(j=0;j<(len>>1)&&i+j<2*n;j++)
			{
				for(k=0;k<n*2;k++)
				{
					t[i+j][k]=1;
				}
			}
		}
	}
	else
	{
		len=(1<<((K+2)>>1));
		for(i=0;i<2*n;i++)
		{
			for(j=0;j<2*n;j++)
			{
				x=((i%len)>=(len>>1));y=((j%len)>=(len>>1));
				if((x+y)&1) t[i][j]=1;
			}
		}
	}
}
int main()
{
	int a,b,i,j,num;
	scanf("%d%d%d",&n,&D1,&D2);
	Solve(D1,t1);
	Solve(D2,t2);
	/*
	for(i=0;i<2*n;i++)
	{
		for(j=0;j<2*n;j++)
		{
			printf("%d ",t2[i][j]);
		}
		putchar('\n');
	}
	return 0;
	*/
	for(a=0;a<2;a++)
	{
		for(b=0;b<2;b++)
		{
			num=0;
			for(i=0;i<2*n;i++)
			{
				for(j=0;j<2*n;j++)
				{
					if(t1[i][j]==a&&t2[i][j]==b) num++;
				}
			}
			if(num>=n*n)
			{
				num=0;
				for(i=0;i<2*n;i++)
				{
					for(j=0;j<2*n;j++)
					{
						if(t1[i][j]==a&&t2[i][j]==b) 
						{
							num++;
							if(num<=n*n) printf("%d %d\n",i,j);
						}
					}
				}
				return 0;
			}
		}
	}
	return 0;
}