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>
using namespace std;
#define pb push_back
#define mp make_pair
vector<pair<int,int> > Move[2];
const int N=605;
int mask[N][N],n;
bool was[N][N][2];
void DFS(int x, int y, int m, int t, int o)
{
	if(x<=0 || y<=0 || x>n*2 || y>n*2 || was[x][y][m]) return;
	mask[x][y]|=t*o;
	was[x][y][m]=1;
	for(auto p:Move[m])
	{
		DFS(x+p.first,y+p.second,m,t^1,o);
	}
}
int cnt[4];
int main()
{
    int i,j,d1,d2;
    scanf("%i %i %i",&n,&d1,&d2);
    for(i=-2*n;i<=2*n;i++)
	{
		for(j=-2*n;j<=2*n;j++)
		{
			if(i*i+j*j==d1)
			{
				Move[0].pb(mp(i,j));
			}
			if(i*i+j*j==d2)
			{
				Move[1].pb(mp(i,j));
				//printf("%i %i\n",i,j);
			}
		}
	}
	for(i=1;i<=2*n;i++) for(j=1;j<=2*n;j++) DFS(i,j,0,1,1);
	for(i=1;i<=2*n;i++) for(j=1;j<=2*n;j++) DFS(i,j,1,1,2);
	for(i=1;i<=2*n;i++) for(j=1;j<=2*n;j++) cnt[mask[i][j]]++;
	//for(i=1;i<=2*n;i++) for(j=1;j<=2*n;j++) printf("%i%c",mask[i][j],j==2*n?'\n':' ');
	int mx=0;
	for(i=1;i<4;i++) if(cnt[i]>cnt[mx]) mx=i;
	int goal=n*n;
	//printf("mx:%i\n",mx);
	for(i=1;i<=2*n;i++) for(j=1;j<=2*n;j++) if(mask[i][j]==mx)
	{
		if(goal>0) printf("%i %i\n",i-1,j-1),goal--;
	}
	return 0;
}