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<cstdio>
#include<algorithm>
using namespace std;

int n,a[605][605];

void work()
{
	int d,l=1;
	scanf("%d",&d);
	for(; d%4==0; d>>=2)
		l<<=1;
	if(~d&1)
	{
		for(int i=1; i<=n; i+=l)
			if(!(i/l&1))
				for(int k=i; k<=min(n,i+l-1); ++k)
					for(int j=1; j<=n; j++)
						a[k][j]++;
	}
	else if(d%4==1)
	{
		for(int i=1; i<=n; i+=l)
			for(int j=1; j<=n; j+=l)
				if((i/l&1)^(j/l&1)^1)
					for(int k=i; k<=min(n,i+l-1); ++k)
						for(int p=j; p<=min(n,j+l-1); ++p)
							a[k][p]++;
	}
	else
	{
		for(int i=1; i<=n; i++)
			for(int j=1; j<=n; j++)
				a[i][j]++;
	}
}

int main()
{
	scanf("%d",&n);
	n<<=1;
	work(); work();
/*	for(int i=1; i<=n; i++)
	{
		for(int j=1; j<=n; j++)
			printf("%d ",a[i][j]);
		puts("");
	}*/
	for(int s=0,i=1; i<=n; i++)
		for(int j=1; j<=n; j++)
			if(a[i][j]==2)
			{
				++s;
				printf("%d %d\n",i-1,j-1);
				if(s*4==n*n)
					return 0;
			}
	return 0;
}