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<iostream>
using namespace std;
int map[800][800],n,d1,d2,z;
void expt(int d)
{
	int now=0;
	while(d%4==0){
		d/=4;//如果整除4，则x,y均为偶数，可以缩点。 
		now++;
	}
	if(d&1){//除二余一，x,y一奇一偶。可以黑白染色 
		for(int i=0;i<2*n;i++)
			for(int j=0;j<2*n;j++)
				if((i>>now)+(j>>now)&1)
				map[i][j]=1;
	}
	else//除四余二，x,y均为奇数 
	{
		for(int i=0;i<2*n;i++)
			for(int j=0;j<2*n;j++)
			if((i>>now)&1)
			map[i][j]=1;
	}
}
int main()
{
	scanf("%d%d%d",&n,&d1,&d2);
	expt(d1);
	expt(d2);
	for(int i=0;i<2*n;i++)
		for(int j=0;j<2*n;j++)
		if(z<n*n && !map[i][j])
		{
			printf("%d %d\n",i,j);
			z++; 
		}
}