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<cstring>
#include<algorithm>
#include<cmath>
#define MN 1005
using namespace std;
int n,tt,s[4];bool col[2][MN][MN],vis[MN][MN];
const int dx[]={-1,-1,1,1};
const int dy[]={-1,1,-1,1};
struct que{int x,y;}q[MN];
void dfs(int x,int y,int c,int typ){
	vis[x][y]=1,col[typ][x][y]=c;
	for(int i=0;i<4;i++)
		for(int j=1;j<=tt;j++){
			int tx=x+dx[i]*q[j].x,ty=y+dy[i]*q[j].y;
			if(tx<0||ty<0||tx>2*n||ty>2*n||vis[tx][ty])continue;
			dfs(tx,ty,!c,typ);	
		}
}
int main(){
	int x,y;scanf("%d%d%d",&n,&x,&y);
	tt=0;for(int i=0;i<=2*n;i++)if(i*i<=x)for(int j=0;j<=2*n;j++)if(i*i+j*j==x)q[++tt].x=i,q[tt].y=j;
	for(int i=0;i<2*n;i++)for(int j=0;j<2*n;j++)if(!vis[i][j])dfs(i,j,0,0);memset(vis,0,sizeof(vis));
	tt=0;for(int i=0;i<=2*n;i++)if(i*i<=y)for(int j=0;j<=2*n;j++)if(i*i+j*j==y)q[++tt].x=i,q[tt].y=j;
	for(int i=0;i<2*n;i++)for(int j=0;j<2*n;j++)if(!vis[i][j])dfs(i,j,0,1);
	for(int i=0;i<2*n;i++)for(int j=0;j<2*n;j++)s[col[0][i][j]*2+col[1][i][j]]++;
	for(int i=0;i<4;i++)if(s[i]>=n*n){
		int cnt=0;
		for(int j=0;j<2*n&&cnt<n*n;j++)for(int k=0;k<2*n&&cnt<n*n;k++)if(col[0][j][k]*2+col[1][j][k]==i)printf("%d %d\n",j,k),cnt++;
		return 0;
	}
}
