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<bits/stdc++.h>
const int N=603;
using namespace std;
int n,d1,d2,vis[N][N],f,now,cnt[16],
tx[N*4],ty[N*4],t;
void dfs(int x,int y,int v){
	if(x<0||x>=n||y<0||y>=n||
	(vis[x][y]>>f&1))return;
	vis[x][y]|=v<<f;
	for(int i=0;i<t;i++)
	dfs(x+tx[i],y+ty[i],v^2);
}
inline void sol(int d){
	t=0;
	for(int i=-n;i<n;i++)
	for(int j=-n;j<n;j++)
	if(i*i+j*j==d)
	tx[t]=i,ty[t++]=j;
	for(int i=0;i<n;i++)
	for(int j=0;j<n;j++)
	dfs(i,j,1);
}
int main(){
	scanf("%d%d%d",&n,&d1,&d2);
	n<<=1;
	sol(d1),f=2,sol(d2);
	for(int i=0,j;i<n;i++)
	for(j=0;j<n;j++)
	cnt[vis[i][j]]++;
	for(f=0;cnt[f]<n*n/4;f++);
	for(int i=0,j;i<n;i++)
	for(j=0;j<n;j++)
	if(now<n*n/4&&vis[i][j]==f)
	printf("%d %d\n",i,j),now++;
}