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>
int N;
bool mp[610][610][2];
int main(){
	scanf("%d",&N);
	for(int t=0;t<2;t++){
		int D;scanf("%d",&D);
		int d=0;
		while(D%4==0)D/=4,d++;
		//printf("d=%d D=%d\n",d,D);
		if(D%4==2){
			for(int x=0;x<N*2;x++)
				for(int y=0;y<N*2;y++)if((x>>d)&1)mp[x][y][t]=1;
		}
		else{
			for(int x=0;x<N*2;x++)
				for(int y=0;y<N*2;y++)if((x>>d)+(y>>d)&1)mp[x][y][t]=1;//,printf("mark %d %d\n",x,y);
		}
	}
	int tot=0;
	for(int i=0;i<N*2&&tot<N*N;i++)
		for(int j=0;j<N*2&&tot<N*N;j++)
			if(!mp[i][j][0]&&!mp[i][j][1])tot++,printf("%d %d\n",i,j); 
}