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>
#define MAXN 600
using namespace std;

int n, m, cnt, d[5], col[MAXN][MAXN][2];
bool vis[MAXN][MAXN][2];

void dfs(int x, int y, int a, int b)
{
	if(x<0 || x>m || y<0 || y>m || vis[x][y][a]) return;
	vis[x][y][a]=true; col[x][y][a]=b;
	for(int i=0; i*i<=d[a]; i++)
	{
		int j=sqrt(d[a]-i*i);
		if(i*i+j*j!=d[a]) continue;
		dfs(x+i, y+j, a, 3-b);
		dfs(x-i, y+j, a, 3-b);
		dfs(x+i, y-j, a, 3-b);
		dfs(x-i, y-j, a, 3-b);
	}
}

void debug()
{
	for(int i=0; i<=m; i++)
	{
		for(int j=0; j<=m; j++) printf("%d ", col[i][j][0]);
		printf("\n");
	}
	printf("\n\n");
	for(int i=0; i<=m; i++)
	{
		for(int j=0; j<=m; j++) printf("%d ", col[i][j][1]);
		printf("\n");
	}
}

int main()
{
	scanf("%d%d%d", &n, &d[0], &d[1]);
	m=n*2-1;
	for(int i=0; i<=m; i++)
	for(int j=0; j<=m; j++)
	{
		if(!vis[i][j][0]) dfs(i, j, 0, 1);
		if(!vis[i][j][1]) dfs(i, j, 1, 1);
	}
	for(int i=0; i<=m; i++)
	for(int j=0; j<=m; j++)
	{
		if(col[i][j][0]==1 && col[i][j][1]==1)
		{
			printf("%d %d \n", i, j);
			if(++cnt==n*n) return 0;
		}
	}
} 