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
//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 20021225
using namespace std;
int read()
{
	int s=0,t=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')	t=-1; ch=getchar();}
	while(ch>='0' && ch<='9')	s=s*10+ch-'0',ch=getchar();
	return s*t;
}
#define N 610
bool vis[N][N]; int top,n;
void work(int dis)
{
	int rem=0; while(!(dis&3))	dis>>=2,rem++;
	if(dis&1)
	{
		for(int i=0;i<top;i++)	for(int j=0;j<top;j++)
			if((i>>rem)+(j>>rem)&1)	vis[i][j]=1;
	}
	else
	{
		for(int i=0;i<top;i++)	if(i>>rem&1)
			for(int j=0;j<top;j++)	vis[i][j]=1;
	}
}
int main()
{
	n=read(); int d1=read(),d2=read(); top=n<<1;
	work(d1); work(d2); int owo=0;
	for(int i=0;i<top;i++)	for(int j=0;j<top;j++)
	{
		if(!vis[i][j])	printf("%d %d\n",i,j),owo++;
		if(owo==n*n)	return 0;
	}
	return 0;
}