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>
using namespace std;
struct data{
	int x,y;
}q1[410001],q2[410001];
int i,j,k,n,m,c[410001],d1,d2,t1,t2,w[111],tmp;
int gv(int x,int y){
	return x*m+y;
}
void dfs1(int x,int y){
	int i,tx,ty,k=c[gv(x,y)];
	for (i=1;i<=t1;i++){
		tx=x+q1[i].x;ty=y+q1[i].y;
		if (tx>=0&&tx<m&&ty>=0&&ty<m&&!c[gv(tx,ty)]) c[gv(tx,ty)]=3-k,dfs1(tx,ty);
	}
}
void dfs2(int x,int y){
	int i,tx,ty,k=c[gv(x,y)]/10;
	for (i=1;i<=t2;i++){
		tx=x+q2[i].x;ty=y+q2[i].y;
		if (tx>=0&&tx<m&&ty>=0&&ty<m&&c[gv(tx,ty)]<10) c[gv(tx,ty)]+=(3-k)*10,dfs2(tx,ty);
	}
}
int main(){
	scanf("%d%d%d",&n,&d1,&d2);m=2*n;memset(c,0,sizeof(c));memset(w,0,sizeof(w));t1=t2=0;
	for (i=0;i<m;i++)
	 for (j=0;j<m;j++){
	 	k=i*i+j*j;
	 	if (k==d1){
	 		q1[++t1].x=i;q1[t1].y=j;
	 		q1[++t1].x=-i;q1[t1].y=j;
	 		q1[++t1].x=i;q1[t1].y=-j;
	 		q1[++t1].x=-i;q1[t1].y=-j;
		}
		if (k==d2){
			q2[++t2].x=i;q2[t2].y=j;
			q2[++t2].x=-i;q2[t2].y=j;
			q2[++t2].x=i;q2[t2].y=-j;
			q2[++t2].x=-i;q2[t2].y=-j;
		}
	 }
	for (i=0;i<m;i++)
	 for (j=0;j<m;j++)
	  if (!c[gv(i,j)]) c[gv(i,j)]=1,dfs1(i,j);
	for (i=0;i<m;i++)
	 for (j=0;j<m;j++)
	  if (c[gv(i,j)]<10) c[gv(i,j)]+=10,dfs2(i,j);
	for (i=0;i<m;i++)
	 for (j=0;j<m;j++) w[c[gv(i,j)]]++;
	for (k=i=1;i<=22;i++)
	 if (w[i]>w[k]) k=i;
	for (i=tmp=0;i<m&&tmp<n*n;i++)
	 for (j=0;j<m&&tmp<n*n;j++)
	  if (c[gv(i,j)]==k) printf("%d %d\n",i,j),tmp++;
	return 0;
}