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 d1,d2,n,f[2001][2001],a=0,now=0;
void color(int x){
	if (x % 2==1){
		for (int i=0;i<2*n;i++)
			for (int j=0;j<2*n;j++)
				if (((i>>now)+(j>>now)) %2) f[i][j]=1;
	}
	else if (x % 4==2){
		for (int i=0;i<2*n;i++)
			for (int j=0;j<2*n;j++)
				if ((i>>now) % 2) f[i][j]=1;
	}
	else{ 
		now++;
		color(x/4);
	} 
}


int main(){
	scanf("%d%d%d",&n,&d1,&d2);
	color(d1);
	now=0;
	color(d2);
	for (int i=0;i<2*n;i++)
		for (int j=0;j<2*n;j++)
			if (f[i][j]==0){
				a++;
				printf("%d %d\n",i,j);
				if (a==n*n) return 0;
			} 
} 