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<stdio.h>
int num[4];
int main()
{
	int n,d1,d2;
	scanf("%d%d%d",&n,&d1,&d2);
	int num1=0,num2=0;
	while((d1&1)==0)
	{
		num1++;
		d1>>=1;
	}
	while((d2&1)==0)
	{
		num2++;
		d2>>=1;
	}
	for(int i=0;i<(n<<1);i++)
	{
		for(int j=0;j<(n<<1);j++)
		{
			int tp1,tp2;
			if(num1&1)
			{
				tp1=((i>>(num1>>1))&1);
			}
			else
			{
				tp1=(((i>>(num1>>1))+(j>>(num1>>1)))&1);
			}
			if(num2&1)
			{
				tp2=((i>>(num2>>1))&1);
			}
			else
			{
				tp2=(((i>>(num2>>1))+(j>>(num2>>1)))&1);
			}
			num[tp1+tp1+tp2]++;
		}
	}
	int wz;
	for(wz=0;wz<4;wz++)
	{
		if(num[wz]>=n*n)
		{
			break;
		}
	}
	int rm=n*n;
	for(int i=0;i<(n<<1);i++)
	{
		for(int j=0;j<(n<<1);j++)
		{
			int tp1,tp2;
			if(num1&1)
			{
				tp1=((i>>(num1>>1))&1);
			}
			else
			{
				tp1=(((i>>(num1>>1))+(j>>(num1>>1)))&1);
			}
			if(num2&1)
			{
				tp2=((i>>(num2>>1))&1);
			}
			else
			{
				tp2=(((i>>(num2>>1))+(j>>(num2>>1)))&1);
			}
			if(tp1+tp1+tp2==wz)
			{
				printf("%d %d\n",i,j);
				rm--;
				if(rm==0)
				{
					return 0;
				}
			}
		}
	}
	return 0;
}