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<vector>
using namespace std;
struct data
{
	int x,y,ox,oy;
};
int n,d1,d2;
bool c1[600][600],c2[600][600];
int cnt[4],tot;
void work(vector<data>&p,int d,bool c[600][600])
{
//	for(auto i:p)printf("%d %d\n",i.ox,i.oy);
//	printf("\n");
	if(d&1)
	{
		for(auto i:p)c[i.ox][i.oy]=(i.x+i.y)&1;
		return ;
	}
	vector<data>q;
	for(auto i:p)if((i.x+i.y)&1)q.push_back(data{(i.x-i.y+1)/2,(i.x+i.y+1)/2,i.ox,i.oy});
	work(q,d/2,c);
	q.clear();
	for(auto i:p)if((i.x+i.y+1)&1)q.push_back(data{(i.x-i.y)/2,(i.x+i.y)/2,i.ox,i.oy});
	work(q,d/2,c);
}
int main()
{
	scanf("%d %d %d",&n,&d1,&d2);
	vector<data>p;
	for(int i=0;i<2*n;++i)
		for(int j=0;j<2*n;++j)
			p.push_back({i,j,i,j});
	work(p,d1,c1);
	work(p,d2,c2);
/*	for(int i=0;i<2*n;++i)
	{
		for(int j=0;j<2*n;++j)printf("%d ",c1[i][j]);
		printf("\n");
	}
	for(int i=0;i<2*n;++i)
	{
		for(int j=0;j<2*n;++j)printf("%d ",c2[i][j]);
		printf("\n");
	}	*/
	for(int i=0;i<2*n;++i)
		for(int j=0;j<2*n;++j)
			++cnt[c1[i][j]<<1|c2[i][j]];
	for(int k=0;k<=3;++k)
		if(cnt[k]>=n*n)
			for(int i=0;i<2*n;++i)
				for(int j=0;j<2*n;++j)
					if((c1[i][j]<<1|c2[i][j])==k)
					{
						++tot;
						printf("%d %d\n",i,j);
						if(tot==n*n)return 0;
					}
	return 0;
}
