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>
const int Maxn=600;
const int D[4][2]={{1,1},{1,-1},{-1,1},{-1,-1}};
int n,d[2];
int step[2][2][Maxn*Maxn+5];
int len[2];
int num[5];
int col[2][Maxn+5][Maxn+5];
void dfs(int k,int x,int y,int c){
	if(x<1||y<1||x>n||y>n||col[k][x][y]!=-1){
		return;
	}
	col[k][x][y]=c;
	for(int i=1;i<=len[k];i++){
		for(int j=0;j<4;j++){
			dfs(k,x+D[j][0]*step[k][0][i],y+D[j][1]*step[k][1][i],c^1);
		}
	}
}
int main(){
	scanf("%d%d%d",&n,&d[0],&d[1]);
	n<<=1;
	for(int k=0;k<2;k++){
		for(int i=0;i<=n;i++){
			for(int j=0;j<=n;j++){
				if(i*i+j*j==d[k]){
					step[k][0][++len[k]]=i;
					step[k][1][len[k]]=j;
				}
			}
		}
	}
	memset(col,-1,sizeof col);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			dfs(0,i,j,0);
			dfs(1,i,j,0);
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			num[col[1][i][j]<<1|col[0][i][j]]++;
		}
	}
	int tmp;
	for(int i=0;i<4;i++){
		if(num[i]>=(n*n>>2)){
			tmp=i;
		}
	}
	int sz=0;
	for(int i=1;i<=n&&(sz<(n*n>>2));i++){
		for(int j=1;j<=n&&(sz<(n*n>>2));j++){
			if((col[1][i][j]<<1|col[0][i][j])==tmp){
				sz++;
				printf("%d %d\n",i-1,j-1);
			}
		}
	}
	return 0;
}
