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>
#define N 600
int a[2][2],i,j,m,n,o,p,x,y;bool b[N][N],c[N][N];
int main()
{
	scanf("%d%d%d",&n,&o,&p),m=n<<1;
	while(~o&1)o>>=1,x++;
	while(~p&1)p>>=1,y++;
	if(x&y&1)for(i=0,x=1<<(x>>1),y=1<<(y>>1);i<m;i++)for(j=0;j<m;j++)a[b[i][j]=i&x][c[i][j]=i&y]++;
	else if(x&1)for(i=0,x=1<<(x>>1),y=1<<(y>>1);i<m;i++)for(j=0;j<m;j++)a[b[i][j]=i&x][c[i][j]=i+j&y]++;
	else if(y&1)for(i=0,x=1<<(x>>1),y=1<<(y>>1);i<m;i++)for(j=0;j<m;j++)a[b[i][j]=i+j&x][c[i][j]=i&y]++;
	else for(i=0,x=1<<(x>>1),y=1<<(y>>1);i<m;i++)for(j=0;j<m;j++)a[b[i][j]=i+j&x][c[i][j]=i+j&y]++;
	if(a[0][0]>=n*n)x=y=0;
	else if(a[0][1]>=n*n)x=0,y=1;
	else if(a[1][0]>=n*n)x=1,y=0;
	else x=y=1;
	for(i=0,n*=n;i<m&&n;i++)for(j=0;j<m&&n;j++)if(b[i][j]==x&&c[i][j]==y)n--,printf("%d %d\n",i,j);
	return 0;
}